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Mock TestsIntroductory Statistics · MST 02
Textbook Workshop
IIFT Kakinada · IPM 2025–30 · Second Semester

Mock Test 1 — Introductory Statistics (MST 02)Exam level

Time: 2 Hours 30 Minutes · Max Marks: 50 · No negative marking
Instructions Answer all eight questions with full steps, in order, without skipping working. Write every final answer correct to four decimal places and underline it. A scientific non-programmable calculator and statistical tables are allowed — the Tables tab has the Z, t and χ² tables you need. There is no negative marking: an attempt with the right method earns method marks even if the arithmetic slips, so never leave a part blank. Where a distribution is not stated, make a reasonable statistical assumption and justify it.
2:30:00
Question 1 — Normal distribution 5 marks
Working backwards from percentages to μ and σ

The daily wages of workers in a large factory are normally distributed. It is found that 10% of the workers earn less than ₹200 per day, while 20% of them earn more than ₹500 per day.

Find the mean and the standard deviation of the daily wage, and hence the percentage of workers earning between ₹300 and ₹400 per day.

Hint: each percentage gives you one z-value. Two z-values, two unknowns.

  1. Turn each percentage into a z. $P(X < 200) = 0.10$, so the area to the left is 0.10 and from the Z-table $z_1 = -1.28$. $P(X > 500) = 0.20$ means $P(X < 500) = 0.80$, so $z_2 = 0.84$.
  2. Write the two standardising equations. $$\frac{200-\mu}{\sigma} = -1.28 \qquad \frac{500-\mu}{\sigma} = 0.84$$
  3. Subtract to kill μ. $\dfrac{500-200}{\sigma} = 0.84-(-1.28) = 2.12$, so $\sigma = \dfrac{300}{2.12} = 141.5094$.
  4. Back-substitute for μ. $\mu = 200 + 1.28(141.5094) = 381.1321$.
  5. Now the band 300 to 400. $z_{300} = \dfrac{300-381.1321}{141.5094} = -0.57$ and $z_{400} = \dfrac{400-381.1321}{141.5094} = 0.13$.
  6. Read the table and subtract. $\Phi(0.13) = 0.5517$, $\Phi(-0.57) = 0.2843$, so $P = 0.5517 - 0.2843 = 0.2674$.
μ = ₹381.1321 · σ = ₹141.5094 · 26.74% of workers earn between ₹300 and ₹400
Question 2 — Chebyshev's inequality 5 marks
Bounding a probability with no distribution assumed

A fair spinner is marked 1, 2, 3, 4, 5 and each number is equally likely. Let $X$ be the number obtained in one spin, with mean $\mu$.

Show that Chebyshev's inequality gives $P\{|X-\mu| \ge 2\} \le 0.5$, and compare this bound with the exact probability. Comment on what the comparison shows.

  1. Mean. $\mu = E(X) = \dfrac{1+2+3+4+5}{5} = 3$.
  2. Variance. $E(X^2) = \dfrac{1+4+9+16+25}{5} = \dfrac{55}{5} = 11$, so $\sigma^2 = E(X^2) - \mu^2 = 11 - 9 = 2$ and $\sigma = \sqrt{2} = 1.4142$.
  3. Apply Chebyshev. In the form $P\{|X-\mu| \ge c\} \le \dfrac{\sigma^2}{c^2}$ with $c = 2$: $$P\{|X-3| \ge 2\} \le \frac{2}{2^2} = \frac{2}{4} = 0.5000$$
  4. Exact probability. $|X-3| \ge 2$ happens only when $X = 1$ or $X = 5$, so the exact probability is $\dfrac{2}{5} = 0.4000$.
  5. Comment. $0.4000 \le 0.5000$, so the inequality holds. The bound is correct but loose — that is the price of using only μ and σ and assuming nothing about the shape of the distribution. Chebyshev never promises a tight answer, only a guaranteed ceiling.
Chebyshev bound = 0.5000 · exact = 0.4000 · the bound holds and is loose
Question 3 — Sampling distribution of $\bar{X}$ 2+2+2 marks
Central Limit Theorem applied to a sample mean

A café serves on average 240 customers per day. Assume the population standard deviation is 45 customers. A random sample of 36 days is taken.

a. What is the probability that the sample mean is more than 252 customers?

b. What is the probability that the sample mean is less than 231 customers?

c. What is the probability that the sample mean is between 234 and 249 customers?

  1. Standard error first — it is used in all three parts. $n = 36 \ge 30$, so by the Central Limit Theorem $\bar{X}$ is approximately normal with $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{45}{\sqrt{36}} = \frac{45}{6} = 7.5$$
  2. (a) $z = \dfrac{252-240}{7.5} = 1.60$. From the table $\Phi(1.60) = 0.9452$, so $P(\bar{X} > 252) = 1 - 0.9452 = 0.0548$.
  3. (b) $z = \dfrac{231-240}{7.5} = -1.20$, giving $P(\bar{X} < 231) = \Phi(-1.20) = 0.1151$.
  4. (c) $z_1 = \dfrac{234-240}{7.5} = -0.80$ and $z_2 = \dfrac{249-240}{7.5} = 1.20$. Then $P = \Phi(1.20) - \Phi(-0.80) = 0.8849 - 0.2119 = 0.6730$.
a. 0.0548 · b. 0.1151 · c. 0.6730
Question 4 — Confidence interval for μ 6 marks
σ unknown, small sample — the t-distribution

Let $\{12,\ 15,\ 11,\ 18,\ 14,\ 16,\ 13,\ 17\}$ be a random sample drawn from a $N(\mu, \sigma^2)$ distribution, where both $\mu$ and $\sigma^2$ are unknown.

Find the 95% confidence interval for $\mu$.

  1. Sample mean. $\sum x = 116$, $n = 8$, so $\bar{x} = \dfrac{116}{8} = 14.5$.
  2. Sample variance. The deviations are $-2.5, 0.5, -3.5, 3.5, -0.5, 1.5, -1.5, 2.5$ and their squares sum to $\sum(x-\bar{x})^2 = 42$. Therefore $$s^2 = \frac{42}{8-1} = 6 \qquad s = \sqrt{6} = 2.4495$$
  3. Choose the tool. σ is unknown and $n = 8$ is small, so use $t$ with $n-1 = 7$ degrees of freedom, not $z$.
  4. Critical value. For 95% two-sided, $t_{0.025,\,7} = 2.365$.
  5. Standard error and margin. $\dfrac{s}{\sqrt{n}} = \dfrac{2.4495}{\sqrt{8}} = 0.8660$, so the margin of error is $2.365 \times 0.8660 = 2.0482$.
  6. Assemble. $14.5 \pm 2.0482$.
95% CI for μ = (12.4518, 16.5482)
Question 5 — Test for a proportion 7 marks
p-value method, one-tailed

An online pharmacy claimed that at least 85% of its orders are delivered within 24 hours. However, a random sample of 250 orders showed that only 198 were delivered within 24 hours.

Test the claim of the pharmacy at 5% level of significance by using the p-value method.

  1. Hypotheses. The claim is the status quo, so it goes in $H_0$. The sample is below the claim, so the alternative points left: $$H_0: p \ge 0.85 \qquad H_1: p < 0.85$$
  2. Sample proportion. $\hat{p} = \dfrac{198}{250} = 0.7920$.
  3. Standard error under $H_0$. Always use the hypothesised $p_0$, never $\hat{p}$: $$\sigma_{\hat{p}} = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.85 \times 0.15}{250}} = \sqrt{0.00051} = 0.0226$$
  4. Test statistic. $z = \dfrac{0.7920 - 0.85}{0.022583} = \dfrac{-0.0580}{0.022583} = -2.5683$, which you look up as $-2.57$.
  5. p-value. Left-tailed, so the p-value is the whole left tail: $P(Z < -2.57) = 0.0051$.
  6. Decide. $0.0051 < 0.05$, so reject $H_0$ at the 5% level.
  7. Conclude in words. There is sufficient evidence at the 5% level that fewer than 85% of orders arrive within 24 hours. The pharmacy's claim is not supported.
z = −2.5683 · p-value = 0.0051 < 0.05 · reject H₀, the claim is not supported
Question 6 — Test for a population variance 6 marks
Chi-square test, two-tailed

The marks obtained by 10 students in a class test are given below.

12151714 16131815 1416

Can we say that the population variance of the distribution of marks, from which the above sample of 10 students was drawn, is equal to 2? Test an appropriate hypothesis at 5% level.

  1. Hypotheses. "Is it equal to 2" is a two-tailed question: $$H_0: \sigma^2 = 2 \qquad H_1: \sigma^2 \ne 2$$
  2. Sample mean. $\sum x = 150$, so $\bar{x} = \dfrac{150}{10} = 15$.
  3. Sum of squared deviations. The deviations are $-3, 0, 2, -1, 1, -2, 3, 0, -1, 1$ and their squares are $9, 0, 4, 1, 1, 4, 9, 0, 1, 1$, so $\sum(x-\bar{x})^2 = 30$ and $s^2 = \dfrac{30}{9} = 3.3333$.
  4. Test statistic. $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{30}{2} = 15.0000 \qquad \text{with } df = 9$$
  5. Critical values. Two-tailed at 5% splits α into 0.025 each side: $\chi^2_{0.975,\,9} = 2.700$ and $\chi^2_{0.025,\,9} = 19.023$.
  6. Decide. $2.700 < 15.0000 < 19.023$, so the statistic falls in the acceptance region. Do not reject $H_0$.
  7. Conclude. At the 5% level there is no evidence against the claim that the population variance of marks equals 2.
χ² = 15.0000, df = 9 · acceptance region (2.700, 19.023) · do not reject H₀
Question 7 — Spearman rank correlation 7 marks
Ranking two variables and measuring agreement

Find the rank correlation coefficient between the hours of revision and the marks scored, from the data given below.

Revision hours 1220825 15301810 2228
Marks scored 41595577 63685046 7284
  1. Rank each row separately. Rank 1 goes to the smallest value. There are no ties, so no correction factor is needed.
  2. Build the working table.
    HoursMarks$R_x$$R_y$ $d$$d^2$
    12413124
    20596511
    85514−39
    257789−11
    156346−24
    306810739
    18505324
    10462200
    227278−11
    2884910−11
    $\sum d^2$34
  3. Apply the formula. With $n = 10$, $n(n^2-1) = 10 \times 99 = 990$: $$r = 1 - \frac{6\sum d^2}{n(n^2-1)} = 1 - \frac{6(34)}{990} = 1 - \frac{204}{990} = 1 - 0.2061 = 0.7939$$
  4. Interpret. A strong positive rank correlation: students who put in more revision hours generally ranked higher in marks, though the agreement is not perfect.
r = 0.7939 (strong positive rank correlation)
Question 8 — Regression and correlation 2+2+2+2 marks
Recovering everything from two regression lines

In a partially destroyed laboratory record of an analysis of bivariate data, the following results only are legible.

Legible $\operatorname{Var}(X) = 25$, and the two regression lines are $$24X - 25Y + 260 = 0 \qquad \text{and} \qquad 8X - 3Y = 20$$

a. The mean values of $X$ and $Y$.

b. Which line is which, with justification.

c. The coefficient of correlation between $X$ and $Y$.

d. The standard deviation of $Y$.

  1. (a) Both regression lines pass through $(\bar{x}, \bar{y})$. So solve them simultaneously. From the second, $8\bar{x} = 3\bar{y} + 20$, i.e. $\bar{x} = \dfrac{3\bar{y}+20}{8}$. Substituting into the first: $$24\left(\frac{3\bar{y}+20}{8}\right) - 25\bar{y} + 260 = 0 \;\Rightarrow\; 3(3\bar{y}+20) - 25\bar{y} + 260 = 0$$ $$9\bar{y} + 60 - 25\bar{y} + 260 = 0 \;\Rightarrow\; -16\bar{y} = -320 \;\Rightarrow\; \bar{y} = 20$$ and then $\bar{x} = \dfrac{3(20)+20}{8} = \dfrac{80}{8} = 10$.
  2. (b) Test the assignment with $r^2 = b_{yx}\,b_{xy} \le 1$. Take the first line as $Y$ on $X$: $25Y = 24X + 260$ gives $b_{yx} = \dfrac{24}{25} = 0.96$. Take the second as $X$ on $Y$: $8X = 3Y + 20$ gives $b_{xy} = \dfrac{3}{8} = 0.375$. Then $r^2 = 0.96 \times 0.375 = 0.36 \le 1$ — valid.
  3. Check the other way round fails. If the first were $X$ on $Y$ then $b_{xy} = \dfrac{25}{24}$, and the second as $Y$ on $X$ gives $b_{yx} = \dfrac{8}{3}$, so $r^2 = \dfrac{25}{24} \times \dfrac{8}{3} = 2.7778 > 1$, which is impossible. The first assignment is therefore the correct one.
  4. (c) Correlation. $r = \pm\sqrt{0.36} = \pm 0.6$. Both slopes are positive, so $r$ takes the positive sign: $r = 0.6000$.
  5. (d) Standard deviation of Y. $\operatorname{Var}(X) = 25$ gives $\sigma_x = 5$. From $b_{yx} = r\dfrac{\sigma_y}{\sigma_x}$: $$0.96 = 0.6 \times \frac{\sigma_y}{5} \;\Rightarrow\; \sigma_y = \frac{0.96 \times 5}{0.6} = 8.0000$$
a. x̄ = 10, ȳ = 20 · b. 24X−25Y+260=0 is Y on X · c. r = 0.6000 · d. σ_y = 8.0000

Paper 1 score

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Finish the paper to unlock the solutions, then award yourself marks question by question. No negative marking: give yourself the method marks you earned even where the final number came out wrong.

IIFT Kakinada · IPM 2025–30 · Second Semester

Mock Test 2 — Introductory Statistics (MST 02)Exam level

Time: 2 Hours 30 Minutes · Max Marks: 50 · No negative marking
Instructions Answer all eight questions with full steps, in order, without skipping working. Write every final answer correct to four decimal places and underline it. A scientific non-programmable calculator and statistical tables are allowed — the Tables tab has the Z, t and χ² tables you need. There is no negative marking: an attempt with the right method earns method marks even if the arithmetic slips, so never leave a part blank. Where a distribution is not stated, make a reasonable statistical assumption and justify it.
2:30:00
Question 1 — Normal distribution 5 marks
Classification into divisions — the end-term's own Q1 pattern

In an examination it is laid down that a student passes if he secures 35 percent or more marks. He is placed in the first, second or third division according as he secures 60% or more marks, between 45% and 60% marks, and marks between 35% and 45% respectively. It is noticed from the result that 20% of the students failed in the examination, whereas 10% of them secured 70% or more marks.

Calculate the percentage of students placed in the first division and in the second division. (Assume the normal distribution of marks.)

Hint: the two facts you are given are the two percentages, not μ and σ. Recover μ and σ first, then every division is just a band.

  1. Translate the two facts into z-values. 20% failed means $P(X < 35) = 0.20$, so $z_1 = -0.84$. And 10% scored 70 or more means $P(X > 70) = 0.10$, i.e. $P(X < 70) = 0.90$, so $z_2 = 1.28$.
  2. Two standardising equations, two unknowns. $$\frac{35-\mu}{\sigma} = -0.84 \qquad \frac{70-\mu}{\sigma} = 1.28$$
  3. Subtract to eliminate μ. $\dfrac{70-35}{\sigma} = 1.28-(-0.84) = 2.12$, so $$\sigma = \frac{35}{2.12} = 16.5094$$
  4. Back-substitute for μ. $\mu = 35 + 0.84(16.5094) = 35 + 13.8679 = 48.8679$.
  5. First division is $X \ge 60$. $z = \dfrac{60-48.8679}{16.5094} = \dfrac{11.1321}{16.5094} = 0.67$, so $$P(X \ge 60) = 1 - \Phi(0.67) = 1 - 0.7486 = 0.2514$$
  6. Second division is $45 \le X < 60$. $z_{45} = \dfrac{45-48.8679}{16.5094} = -0.23$, and $\Phi(-0.23) = 0.4090$. So $$P(45 \le X < 60) = 0.7486 - 0.4090 = 0.3396$$
  7. Sense-check. Failed 20%, third division $= 0.4090-0.2000 = 20.90\%$, second 33.96%, first 25.14% — these sum to 100%, as they must.
μ = 48.8679, σ = 16.5094 · first division = 25.14% · second division = 33.96%
Question 2 — Uniform distribution and Chebyshev 5 marks
A bound that ignores the shape, against the truth

The waiting time $X$ (in minutes) for a shuttle bus is uniformly distributed over the interval $[0, 10]$.

Using Chebyshev's inequality, find an upper bound for $P\{|X-\mu| \ge 4\}$. Compute the exact probability from the uniform distribution and comment on the difference.

  1. Mean and variance of a uniform. For $X \sim U(a,b)$ with $a = 0$, $b = 10$: $$\mu = \frac{a+b}{2} = 5 \qquad \sigma^2 = \frac{(b-a)^2}{12} = \frac{100}{12} = 8.3333$$ so $\sigma = 2.8868$ minutes.
  2. Apply Chebyshev. With $c = 4$: $$P\{|X-5| \ge 4\} \le \frac{\sigma^2}{c^2} = \frac{8.3333}{16} = 0.5208$$ Equivalently $c = k\sigma$ gives $k = \dfrac{4}{2.8868} = 1.3856$ and $\dfrac{1}{k^2} = 0.5208$.
  3. Exact probability. $|X-5| \ge 4$ means $X \le 1$ or $X \ge 9$. For a uniform on $[0,10]$ the density is $\dfrac{1}{10}$, so $$P = \frac{1-0}{10} + \frac{10-9}{10} = 0.1 + 0.1 = 0.2000$$
  4. Comment. The true probability 0.2000 is far below the Chebyshev ceiling 0.5208. The bound is valid but very loose here, because Chebyshev must cover every distribution with this mean and variance, including heavy-tailed ones. The uniform has no tails at all beyond its endpoints, so it sits comfortably inside the guarantee.
Chebyshev bound = 0.5208 · exact = 0.2000 · the bound holds and is loose
Question 3 — Sampling distribution of $\bar{X}$ 2+2+2 marks
Central Limit Theorem applied to a sample mean

According to a manufacturer, a certain brand of battery lasts on average 500 hours. Assume the population standard deviation is 64 hours. Suppose a random sample of 64 batteries is taken.

a. What is the probability that the sample mean is more than 512 hours?

b. What is the probability that the sample mean is less than 494 hours?

c. What is the probability that the sample mean is between 490 and 516 hours?

  1. Standard error first. $n = 64 \ge 30$, so by the Central Limit Theorem $\bar{X}$ is approximately normal with $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{64}{\sqrt{64}} = \frac{64}{8} = 8$$
  2. (a) $z = \dfrac{512-500}{8} = 1.50$. From the table $\Phi(1.50) = 0.9332$, so $P(\bar{X} > 512) = 1 - 0.9332 = 0.0668$.
  3. (b) $z = \dfrac{494-500}{8} = -0.75$, giving $P(\bar{X} < 494) = \Phi(-0.75) = 0.2266$.
  4. (c) $z_1 = \dfrac{490-500}{8} = -1.25$ and $z_2 = \dfrac{516-500}{8} = 2.00$. Then $P = \Phi(2.00) - \Phi(-1.25) = 0.9772 - 0.1056 = 0.8716$.
a. 0.0668 · b. 0.2266 · c. 0.8716
Question 4 — Confidence interval for μ 6 marks
σ unknown, small sample — the t-distribution

Let $\{2,\ 4,\ 6,\ 8,\ 10,\ 12,\ 14\}$ be the set of collected random sample values from a $N(\mu, \sigma^2)$ distribution, where both $\mu$ and $\sigma^2$ are unknown.

Find the 95% confidence interval for $\mu$.

  1. Sample mean. $\sum x = 56$, $n = 7$, so $\bar{x} = \dfrac{56}{7} = 8$.
  2. Sample variance. The deviations are $-6, -4, -2, 0, 2, 4, 6$ and their squares are $36, 16, 4, 0, 4, 16, 36$, summing to $\sum(x-\bar{x})^2 = 112$. Therefore $$s^2 = \frac{112}{7-1} = 18.6667 \qquad s = \sqrt{18.6667} = 4.3205$$
  3. Choose the tool. σ is unknown and $n = 7$ is small, so use $t$ with $n-1 = 6$ degrees of freedom.
  4. Critical value. For 95% two-sided, $t_{0.025,\,6} = 2.447$.
  5. Standard error and margin. $\dfrac{s}{\sqrt{n}} = \dfrac{4.3205}{\sqrt{7}} = 1.6330$, so the margin of error is $2.447 \times 1.6330 = 3.9959$.
  6. Assemble. $8 \pm 3.9959$.
95% CI for μ = (4.0041, 11.9959)
Question 5 — Test for a proportion 7 marks
p-value method, one-tailed

A political candidate claimed that more than 60% of the voters in a constituency support her. A random sample of 400 voters showed that 260 of them support her.

Test the claim of the candidate at 5% level of significance by using the p-value method.

  1. Hypotheses. The claim contains "more than", which has no equality in it, so the claim becomes the alternative: $$H_0: p \le 0.60 \qquad H_1: p > 0.60$$
  2. Sample proportion. $\hat{p} = \dfrac{260}{400} = 0.6500$.
  3. Standard error under $H_0$. Use the hypothesised $p_0 = 0.60$: $$\sigma_{\hat{p}} = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.60 \times 0.40}{400}} = \sqrt{0.0006} = 0.0245$$
  4. Test statistic. $z = \dfrac{0.6500 - 0.60}{0.0245} = \dfrac{0.0500}{0.0245} = 2.04$.
  5. p-value. Right-tailed, so the p-value is the right tail: $P(Z > 2.04) = 1 - 0.9793 = 0.0207$.
  6. Decide. $0.0207 < 0.05$, so reject $H_0$ at the 5% level.
  7. Conclude in words. There is sufficient evidence at the 5% level that more than 60% of voters support the candidate. The claim is supported by the sample.
z = 2.0412 · p-value = 0.0207 < 0.05 · reject H₀, the claim is supported
Question 6 — Test for a population variance 6 marks
Chi-square test, two-tailed

The fill volumes (in millilitres, measured as deviations from the nominal value) of 8 randomly chosen bottles are given below.

25283027 32292631

Can we say that the population variance of the distribution of fill volumes, from which the above sample of 8 bottles was drawn, is equal to 2 square millilitres? Test an appropriate hypothesis at 5% level.

  1. Hypotheses. Two-tailed, because the question asks whether it equals 2: $$H_0: \sigma^2 = 2 \qquad H_1: \sigma^2 \ne 2$$
  2. Sample mean. $\sum x = 228$, so $\bar{x} = \dfrac{228}{8} = 28.5$.
  3. Sum of squared deviations. The deviations are $-3.5, -0.5, 1.5, -1.5, 3.5, 0.5, -2.5, 2.5$ and their squares are $12.25, 0.25, 2.25, 2.25, 12.25, 0.25, 6.25, 6.25$, so $\sum(x-\bar{x})^2 = 42$ and $s^2 = \dfrac{42}{7} = 6$.
  4. Test statistic. $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{42}{2} = 21.0000 \qquad \text{with } df = 7$$
  5. Critical values. Two-tailed at 5%: $\chi^2_{0.975,\,7} = 1.690$ and $\chi^2_{0.025,\,7} = 16.013$.
  6. Decide. $21.0000 > 16.013$, so the statistic falls in the right rejection region. Reject $H_0$.
  7. Conclude. At the 5% level there is sufficient evidence that the population variance of fill volumes is not equal to 2 square millilitres; the sample suggests it is larger.
χ² = 21.0000, df = 7 · 21.0000 > 16.013 · reject H₀
Question 7 — Spearman rank correlation 7 marks
Ranking two variables and measuring agreement

Find the rank correlation coefficient between the price and the demand from the data given below.

Price (Rs.) 10152025 30354045 5055
Demand (units) 95889180 75786570 5855
  1. Rank each row separately. Rank 1 goes to the smallest value. There are no ties, so no correction factor is needed. Price is already in increasing order, so $R_x = 1$ to $10$.
  2. Build the working table.
    PriceDemand$R_x$$R_y$ $d$$d^2$
    1095110−981
    158828−636
    209139−636
    258047−39
    30755500
    35786600
    406573416
    457084416
    505892749
    5555101981
    $\sum d^2$324
  3. Apply the formula. With $n = 10$, $n(n^2-1) = 990$: $$r = 1 - \frac{6\sum d^2}{n(n^2-1)} = 1 - \frac{6(324)}{990} = 1 - \frac{1944}{990} = 1 - 1.9636 = -0.9636$$
  4. Interpret. A very strong negative rank correlation, exactly what economics predicts: as price rises through the ranks, demand falls through the ranks almost perfectly.
r = −0.9636 (very strong negative rank correlation)
Question 8 — Regression and correlation 2+2+2+2 marks
Recovering everything from two regression lines

In a partially destroyed laboratory record of an analysis of bivariate data, the following results only are legible.

Legible $\operatorname{Var}(X) = 16$, and the two regression lines are $$6X + 5Y - 240 = 0 \qquad \text{and} \qquad 15X + 8Y - 465 = 0$$

a. The mean values of $X$ and $Y$.

b. Which line is which, with justification.

c. The coefficient of correlation between $X$ and $Y$.

d. The standard deviation of $Y$.

  1. (a) Both lines pass through $(\bar{x}, \bar{y})$. Solve $6\bar{x} + 5\bar{y} = 240$ and $15\bar{x} + 8\bar{y} = 465$ simultaneously. Multiply the first by 15 and the second by 6: $$90\bar{x} + 75\bar{y} = 3600 \qquad 90\bar{x} + 48\bar{y} = 2790$$ Subtracting, $27\bar{y} = 810$, so $\bar{y} = 30$. Then $6\bar{x} = 240 - 5(30) = 90$, giving $\bar{x} = 15$.
  2. (b) Test the assignment with $r^2 = b_{yx}\,b_{xy} \le 1$. Take the first line as $Y$ on $X$: $5Y = 240 - 6X$ gives $b_{yx} = -\dfrac{6}{5} = -1.2$. Take the second as $X$ on $Y$: $15X = 465 - 8Y$ gives $b_{xy} = -\dfrac{8}{15} = -0.5333$. Then $r^2 = (-1.2)(-0.5333) = 0.64 \le 1$ — valid.
  3. Check the other way round fails. If the first were $X$ on $Y$ then $b_{xy} = -\dfrac{5}{6}$, and the second as $Y$ on $X$ gives $b_{yx} = -\dfrac{15}{8}$, so $r^2 = \dfrac{5}{6} \times \dfrac{15}{8} = 1.5625 > 1$, which is impossible.
  4. (c) Correlation. $r = \pm\sqrt{0.64} = \pm 0.8$. Both slopes are negative, so $r$ takes the negative sign: $r = -0.8000$.
  5. (d) Standard deviation of Y. $\operatorname{Var}(X) = 16$ gives $\sigma_x = 4$. From $b_{yx} = r\dfrac{\sigma_y}{\sigma_x}$: $$-1.2 = -0.8 \times \frac{\sigma_y}{4} \;\Rightarrow\; \sigma_y = \frac{-1.2 \times 4}{-0.8} = 6.0000$$
a. x̄ = 15, ȳ = 30 · b. 6X+5Y−240=0 is Y on X · c. r = −0.8000 · d. σ_y = 6.0000

Paper 2 score

— / 50

Finish the paper to unlock the solutions, then award yourself marks question by question. No negative marking: give yourself the method marks you earned even where the final number came out wrong.

IIFT Kakinada · IPM 2025–30 · Second Semester

Mock Test 3 — Introductory Statistics (MST 02)Stretch level

Time: 2 Hours 30 Minutes · Max Marks: 50 · No negative marking
Read this first This paper is deliberately harder than the end-term. Every question uses the same eight topics and the same tables, but each one adds a second step the real paper stopped short of — a conditional probability, a sample-size calculation, a Type II error, a p-value squeezed between two table columns, tied ranks. If you can clear 35 here, the real paper will feel slow. Marks are identical to the real paper and there is no negative marking.
2:30:00
Question 1 — Normal distribution 2+1+2 marks
μ and σ recovered, then a guarantee time and a conditional probability

The lifetime of an electronic component is normally distributed. It is found that 10% of the components last less than 1200 hours, while 5% of them last more than 1800 hours.

a. Find the mean and the standard deviation of the lifetime.

b. The manufacturer replaces free of charge any component that fails before a stated guarantee time. If at most 3% of components are to be replaced, what guarantee time should be stated?

c. Among the components that are still working at 1500 hours, what fraction will survive beyond 1800 hours?

Hint for (c): "among those that survived 1500" is the language of conditional probability.

  1. (a) Turn each percentage into a z. $P(X < 1200) = 0.10$ gives $z_1 = -1.28$, and $P(X > 1800) = 0.05$ gives $z_2 = 1.645$.
  2. Two equations, two unknowns. $$\frac{1200-\mu}{\sigma} = -1.28 \qquad \frac{1800-\mu}{\sigma} = 1.645$$ Subtracting, $\dfrac{600}{\sigma} = 2.925$, so $\sigma = 205.1282$ hours, and $\mu = 1200 + 1.28(205.1282) = 1462.5641$ hours.
  3. (b) Guarantee time. At most 3% replaced means the guarantee time $t$ has 3% of the area to its left, so $z = -1.88$: $$t = \mu + z\sigma = 1462.5641 - 1.88(205.1282) = 1462.5641 - 385.6410 = 1076.9231$$ Rounding down to be safe, state a guarantee of about 1076 hours.
  4. (c) Set up the conditional probability. $$P(X > 1800 \mid X > 1500) = \frac{P(X > 1800 \ \cap\ X > 1500)}{P(X > 1500)} = \frac{P(X > 1800)}{P(X > 1500)}$$ because surviving past 1800 already implies surviving past 1500.
  5. Numerator is given. $P(X > 1800) = 0.05$.
  6. Denominator. $z = \dfrac{1500-1462.5641}{205.1282} = 0.18$, so $\Phi(0.18) = 0.5714$ and $P(X > 1500) = 1 - 0.5714 = 0.4286$.
  7. Divide. $\dfrac{0.05}{0.4286} = 0.1167$.
a. μ = 1462.5641 h, σ = 205.1282 h · b. guarantee ≈ 1076.9231 h · c. 0.1167
Question 2 — Markov and Chebyshev compared 1+1+1+2 marks
Two bounds on the same event, and inverting one of them

The daily demand $X$ at a retail counter is a non-negative random variable with mean 60 units and variance 25 units². Nothing else is known about its distribution.

a. Use Markov's inequality to find an upper bound for $P(X \ge 80)$.

b. Use Chebyshev's inequality to find an upper bound for $P(X \ge 80)$.

c. Which bound is sharper, and why?

d. Find the smallest $c$ for which Chebyshev's inequality guarantees $P\{|X-\mu| < c\} \ge 0.96$.

  1. (a) Markov. For a non-negative $X$, $P(X \ge a) \le \dfrac{E(X)}{a}$: $$P(X \ge 80) \le \frac{60}{80} = 0.7500$$
  2. (b) Chebyshev. The event $X \ge 80$ sits inside the two-sided event $|X-60| \ge 20$, so $$P(X \ge 80) \le P\{|X-60| \ge 20\} \le \frac{\sigma^2}{20^2} = \frac{25}{400} = 0.0625$$
  3. (c) Which is sharper. Chebyshev, by a factor of twelve. Markov uses only the mean and so must allow for distributions that put a lot of probability far above 60. Chebyshev also knows the variance is only 25, i.e. $\sigma = 5$, and 80 is four standard deviations above the mean — information Markov simply cannot see. More information, tighter bound.
  4. (d) Invert the inequality. Chebyshev in complement form says $$P\{|X-\mu| < c\} \ge 1 - \frac{\sigma^2}{c^2}$$ so we need $1 - \dfrac{25}{c^2} \ge 0.96$.
  5. Solve. $\dfrac{25}{c^2} \le 0.04 \Rightarrow c^2 \ge \dfrac{25}{0.04} = 625 \Rightarrow c \ge 25$. The smallest such $c$ is 25 units.
  6. Sense-check. $c = 25$ is $k = 5$ standard deviations, and $1 - 1/k^2 = 1 - 1/25 = 0.96$ exactly, as required.
a. 0.7500 · b. 0.0625 · c. Chebyshev, it uses the variance · d. c = 25.0000
Question 3 — Sampling distribution of $\bar{X}$ 2+2+2 marks
A probability, a required sample size, and a finite population

A population has mean 75 and standard deviation 12.

a. For a random sample of 36, find the probability that the sample mean lies within 3 units of the population mean.

b. How large must the sample be so that the sample mean lies within 2 units of the population mean with probability 0.95?

c. Suppose instead the population contains only $N = 200$ members and a sample of $n = 50$ is drawn without replacement. Using the finite population correction $\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}}$, find $P(\bar{X} > 78)$.

  1. (a) Standard error. $\sigma_{\bar{x}} = \dfrac{12}{\sqrt{36}} = 2$. "Within 3 units" means $|\bar{X}-\mu| < 3$, i.e. $|z| < \dfrac{3}{2} = 1.50$.
  2. Two-sided area. $$P(|Z| < 1.50) = 2\Phi(1.50) - 1 = 2(0.9332) - 1 = 0.8664$$
  3. (b) Set the margin and solve for n. For probability 0.95 the z-value is 1.96, and the half-width must equal 2: $$1.96 \times \frac{12}{\sqrt{n}} = 2 \;\Rightarrow\; \sqrt{n} = \frac{1.96 \times 12}{2} = 11.76$$
  4. Square and round up. $n = (11.76)^2 = 138.2976$. Sample size must be a whole number and must not fall short, so round up: $n = 139$.
  5. (c) Apply the correction factor. $$\sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{200-50}{199}} = \sqrt{0.7538} = 0.8682$$ $$\sigma_{\bar{x}} = \frac{12}{\sqrt{50}} \times 0.8682 = 1.6971 \times 0.8682 = 1.4734$$
  6. Standardise and read. $z = \dfrac{78-75}{1.4734} = 2.04$, so $P(\bar{X} > 78) = 1 - \Phi(2.04) = 1 - 0.9793 = 0.0207$.
  7. Note why the correction matters. Without it the standard error would be 1.6971 and $z = 1.77$, giving 0.0384 — nearly double. Sampling a quarter of a small population really does make $\bar{X}$ less variable.
a. 0.8664 · b. n = 139 · c. 0.0207
Question 4 — Confidence intervals 2+2+2 marks
An interval for μ, an interval for σ², and a sample-size consequence

A random sample of 12 observations drawn from a normal population gives $\sum x = 300$ and $\sum x^2 = 7600$.

a. Construct a 99% confidence interval for $\mu$.

b. Construct a 95% confidence interval for $\sigma^2$.

c. Roughly what sample size would be needed to halve the width of the interval in part (a)?

  1. Mean and corrected sum of squares. $\bar{x} = \dfrac{300}{12} = 25$, and using the shortcut $$\sum(x-\bar{x})^2 = \sum x^2 - n\bar{x}^2 = 7600 - 12(25)^2 = 7600 - 7500 = 100$$ so $s^2 = \dfrac{100}{11} = 9.0909$ and $s = 3.0151$.
  2. (a) σ is unknown, so use t with $df = 11$. For 99% two-sided, $t_{0.005,\,11} = 3.106$.
  3. Margin of error. $\dfrac{s}{\sqrt{n}} = \dfrac{3.0151}{\sqrt{12}} = 0.8704$, so the margin is $3.106 \times 0.8704 = 2.7034$, giving $25 \pm 2.7034$.
  4. (b) Interval for a variance uses chi-square, and the tails swap sides. $$\left(\frac{(n-1)s^2}{\chi^2_{0.025,\,11}},\ \frac{(n-1)s^2}{\chi^2_{0.975,\,11}}\right)$$ with $\chi^2_{0.025,\,11} = 21.920$ and $\chi^2_{0.975,\,11} = 3.816$.
  5. Substitute. $(n-1)s^2 = 100$, so the interval is $\left(\dfrac{100}{21.920},\ \dfrac{100}{3.816}\right) = (4.5620,\ 26.2055)$. It is wide and strongly asymmetric — that is normal for a variance interval from only 12 observations.
  6. (c) Width is proportional to $1/\sqrt{n}$. To halve the width you need $\sqrt{n}$ to double, so $n$ must quadruple: $n \approx 4 \times 12 = 48$. (Slightly fewer in truth, because $t$ shrinks towards $z$ as $df$ grows, but 48 is the answer expected.)
a. (22.2966, 27.7034) · b. (4.5620, 26.2055) · c. n ≈ 48
Question 5 — Proportion test and Type II error 4+3 marks
p-value method, then the probability of missing a real shift

A manufacturer claims that at most 5% of the items produced by a machine are defective. In a random sample of 300 items, 22 were found defective.

a. Test the claim at 1% level of significance using the p-value method.

b. Using the same test, find the probability of a Type II error if the true defective rate is in fact 8%.

  1. (a) Hypotheses. "At most 5%" contains equality, so it is the null; the sample is above it, so the alternative points right: $$H_0: p \le 0.05 \qquad H_1: p > 0.05$$
  2. Sample proportion and standard error. $\hat{p} = \dfrac{22}{300} = 0.0733$, and under $H_0$ $$\sigma_{\hat{p}} = \sqrt{\frac{0.05 \times 0.95}{300}} = \sqrt{0.00015833} = 0.0126$$
  3. Test statistic and p-value. $z = \dfrac{0.0733 - 0.05}{0.0126} = 1.85$, so the right-tail p-value is $1 - \Phi(1.85) = 1 - 0.9678 = 0.0322$.
  4. Decide. $0.0322 > 0.01$, so do not reject $H_0$ at the 1% level. There is not enough evidence at 1% to say the defective rate exceeds 5%. (Note it would have been rejected at 5% — the significance level genuinely decides this one.)
  5. (b) Find the decision boundary in $\hat{p}$ units. A Type II error means failing to reject when $H_1$ is true, so first fix the critical value. At $\alpha = 0.01$ right-tailed, $z_{0.01} = 2.33$: $$\hat{p}_{crit} = 0.05 + 2.33(0.012583) = 0.05 + 0.029319 = 0.0793$$ We reject only if $\hat{p} > 0.0793$.
  6. Re-centre on the true value $p = 0.08$. The standard error changes because $p$ changed: $$\sigma_{\hat{p}}' = \sqrt{\frac{0.08 \times 0.92}{300}} = \sqrt{0.00024533} = 0.0157$$
  7. β is the probability of landing in the acceptance region. $$\beta = P(\hat{p} < 0.0793 \mid p = 0.08) = \Phi\!\left(\frac{0.0793 - 0.08}{0.015663}\right) = \Phi(-0.04) = 0.4840$$
  8. Interpret. Even though the true rate is 8%, this test misses it about 48% of the time. The power is only $1 - \beta = 0.5160$. A 1% significance level with $n = 300$ is simply not sensitive enough to catch a shift this small.
a. z = 1.8543, p-value = 0.0322 > 0.01, do not reject H₀ · b. β = 0.4840 (power = 0.5160)
Question 6 — Variance test, one-tailed 4+2 marks
Chi-square in one tail, and squeezing a p-value out of a table

A bottling process is regarded as being in control if the variance of the fill weights does not exceed 1.5 g². A random sample of 16 bottles gives a sample variance of 2.8 g².

a. Test at 5% level whether the process variance exceeds 1.5 g².

b. Using the chi-square table only, place the p-value between two bounds.

  1. (a) Hypotheses. "Does not exceed" contains equality and is the in-control state: $$H_0: \sigma^2 \le 1.5 \qquad H_1: \sigma^2 > 1.5$$ This is one-tailed to the right, so the whole 5% sits in the upper tail.
  2. Test statistic. $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{15 \times 2.8}{1.5} = \frac{42}{1.5} = 28.0000 \qquad df = 15$$
  3. Critical value. $\chi^2_{0.05,\,15} = 24.996$.
  4. Decide. $28.0000 > 24.996$, so reject $H_0$ at the 5% level. There is sufficient evidence that the process variance exceeds 1.5 g² — the process is out of control on consistency.
  5. (b) Locate 28.0000 in the $df = 15$ row. Reading along that row: $\chi^2_{0.025,\,15} = 27.488$ and $\chi^2_{0.010,\,15} = 30.578$.
  6. Sandwich it. Since $27.488 < 28.0000 < 30.578$, the upper-tail area lies between 0.010 and 0.025: $$0.010 < p\text{-value} < 0.025$$ This is consistent with part (a): the whole interval sits below 0.05, so rejection at 5% was never in doubt. It would also be rejected at 2.5%, but not at 1%.
a. χ² = 28.0000 > 24.996, reject H₀ · b. 0.010 < p-value < 0.025
Question 7 — Spearman rank correlation with ties 7 marks
Repeated values, mid-ranks, and the correction factor

Find the rank correlation coefficient between the marks in Statistics and the marks in Economics from the data given below.

Statistics 68726875 80728568 7590
Economics 62706572 78688260 7488

Hint: look at the Statistics row before you start ranking.

  1. Spot the ties. In Statistics, 68 occurs three times, 72 occurs twice and 75 occurs twice. Economics has no repeats. Tied values must share the average of the ranks they would have occupied.
  2. Assign mid-ranks in Statistics. Sorted, the values are $68, 68, 68, 72, 72, 75, 75, 80, 85, 90$ occupying positions 1 to 10. The three 68s occupy positions 1, 2, 3, so each gets $\dfrac{1+2+3}{3} = 2$. The two 72s occupy 4 and 5, so each gets 4.5. The two 75s occupy 6 and 7, so each gets 6.5. Then $80 \to 8$, $85 \to 9$, $90 \to 10$.
  3. Build the working table.
    StatsEcon$R_x$$R_y$ $d$$d^2$
    68622200
    72704.55−0.50.25
    686523−11
    75726.560.50.25
    80788800
    72684.540.50.25
    85829900
    68602111
    75746.57−0.50.25
    9088101000
    $\sum d^2$3
  4. Correction factor, one term per tied group. With $C.F. = \dfrac{m(m^2-1)}{12}$: $$\frac{3(9-1)}{12} = 2 \qquad \frac{2(4-1)}{12} = 0.5 \qquad \frac{2(4-1)}{12} = 0.5$$ so $\sum C.F. = 2 + 0.5 + 0.5 = 3$.
  5. Apply the corrected formula. $$r = 1 - \frac{6}{n(n^2-1)}\left[\sum d_i^2 + \sum \frac{m(m^2-1)}{12}\right] = 1 - \frac{6}{990}\left[3 + 3\right]$$ $$= 1 - \frac{36}{990} = 1 - 0.0364 = 0.9636$$
  6. Interpret. A very strong positive rank correlation — students who rank high in Statistics rank high in Economics almost without exception.
Σd² = 3 · ΣC.F. = 3 · r = 0.9636
Question 8 — Regression, correlation and prediction 2+2+2+2 marks
Identify the lines, recover σ, then predict with the right one

In a partially destroyed laboratory record of an analysis of bivariate data, the following results only are legible.

Legible $\operatorname{Var}(Y) = 4$, and the two regression lines are $$X + 4Y - 32 = 0 \qquad \text{and} \qquad X + Y - 11 = 0$$

a. The mean values of $X$ and $Y$.

b. Identify which line is which, and find the coefficient of correlation.

c. The standard deviation of $X$.

d. Estimate $Y$ when $X = 5$, stating clearly which line you must use and why.

  1. (a) Both lines pass through $(\bar{x}, \bar{y})$. Solve $\bar{x} + 4\bar{y} = 32$ and $\bar{x} + \bar{y} = 11$. Subtracting the second from the first gives $3\bar{y} = 21$, so $\bar{y} = 7$, and then $\bar{x} = 11 - 7 = 4$.
  2. (b) Try the assignment and test $r^2 = b_{yx}b_{xy} \le 1$. Take $X + 4Y - 32 = 0$ as $Y$ on $X$: $4Y = 32 - X$ gives $b_{yx} = -\dfrac{1}{4} = -0.25$. Take $X + Y - 11 = 0$ as $X$ on $Y$: $X = 11 - Y$ gives $b_{xy} = -1$. Then $r^2 = (-0.25)(-1) = 0.25 \le 1$ — valid.
  3. Confirm the alternative is impossible. If $X + 4Y - 32 = 0$ were $X$ on $Y$ then $b_{xy} = -4$, and $X + Y - 11 = 0$ as $Y$ on $X$ gives $b_{yx} = -1$, so $r^2 = 4 > 1$. Rejected. The first assignment stands.
  4. Correlation. $r = \pm\sqrt{0.25} = \pm 0.5$, and both slopes are negative, so $r = -0.5000$.
  5. (c) Standard deviation of X. $\operatorname{Var}(Y) = 4$ gives $\sigma_y = 2$. From $b_{yx} = r\dfrac{\sigma_y}{\sigma_x}$: $$-0.25 = -0.5 \times \frac{2}{\sigma_x} \;\Rightarrow\; -0.25 = \frac{-1}{\sigma_x} \;\Rightarrow\; \sigma_x = 4.0000$$
  6. (d) Choose the line by what is being predicted. We are given $X$ and want $Y$, so we must use the regression of $Y$ on $X$, which is $X + 4Y - 32 = 0$. The other line minimises horizontal deviations and would give a biased answer here.
  7. Substitute. $Y = \dfrac{32 - X}{4} = \dfrac{32 - 5}{4} = \dfrac{27}{4} = 6.7500$.
a. x̄ = 4, ȳ = 7 · b. X+4Y−32=0 is Y on X, r = −0.5000 · c. σ_x = 4.0000 · d. Ŷ = 6.7500

Paper 3 score

— / 50

This paper is harder than the end-term on purpose. 35 or more here is a strong signal you are ready. No negative marking: award yourself the method marks you earned.

IIFT Kakinada · IPM 2025–30 · Second Semester

Mock Test 4 — Introductory Statistics (MST 02)Predicted paper

Time: 2 Hours 30 Minutes · Max Marks: 50 · No negative marking
What this paper is This is the best guess at the actual paper, built by matching the professor's own end-term question for question, then applying his stated split: half the marks easy to moderate, half hard. Questions 1–4 and part (a) of Question 5 are the 26 gettable marks — bank every one of them before you touch the rest. Questions 5(b) to 8 are the 24 hard marks, and each one is hard in the specific way this examiner makes things hard: a second step bolted onto a familiar first.
Strategy for this paper Spend the first 60 minutes on Q1–Q4 and Q5(a). If they are done and checked, you already have 26 marks with 90 minutes left for the remaining 24. That is the whole game — the hard half is where time disappears, so do not enter it with unbanked easy marks behind you.
2:30:00
Question 1 — Normal distribution 5 marks
Moderate Classification into divisions — his exact Q1 pattern

In an examination it is laid down that a student passes if he secures 40 percent or more marks. He is placed in the first, second or third division according as he secures 60% or more marks, between 45% and 60% marks, and marks between 40% and 45% respectively. He gets distinction in case he secures 75% or more marks. It is noticed from the result that 10% of the students failed in the examination, whereas 5% of them obtained distinction.

Calculate the percentage of students placed in the second division. (Assume the normal distribution of marks.)

Hint: the failure rate and the distinction rate are your two equations. Nothing can be computed until μ and σ are out.

  1. Turn the two given percentages into z-values. Failing means scoring under 40, so $P(X < 40) = 0.10$ and $z_1 = -1.28$. Distinction means 75 or more, so $P(X > 75) = 0.05$, i.e. $P(X < 75) = 0.95$ and $z_2 = 1.645$.
  2. Write both standardising equations. $$\frac{40-\mu}{\sigma} = -1.28 \qquad \frac{75-\mu}{\sigma} = 1.645$$
  3. Subtract to eliminate μ. $$\frac{75-40}{\sigma} = 1.645 - (-1.28) = 2.925 \;\Rightarrow\; \sigma = \frac{35}{2.925} = 11.9658$$
  4. Back-substitute for μ. $\mu = 40 + 1.28(11.9658) = 40 + 15.3162 = 55.3162$.
  5. Second division is the band $45 \le X < 60$. $$z_{45} = \frac{45-55.3162}{11.9658} = -0.86 \qquad z_{60} = \frac{60-55.3162}{11.9658} = 0.39$$
  6. Read the table and subtract. $\Phi(0.39) = 0.6517$ and $\Phi(-0.86) = 0.1949$, so $$P(45 \le X < 60) = 0.6517 - 0.1949 = 0.4568$$
  7. Sense-check with the other divisions. First division $= 1 - 0.6517 = 0.3483$, third $= 0.1949 - 0.1000 = 0.0949$, failed $= 0.1000$. Adding: $0.3483 + 0.4568 + 0.0949 + 0.1000 = 1.0000$ ✓
μ = 55.3162, σ = 11.9658 · 45.68% of students are placed in the second division
Question 2 — Chebyshev's inequality 5 marks
Moderate A "show that" proof, exactly as he set it

If $X$ is the random variable denoting the number scored in a throw of a fair die, show that Chebyshev's inequality gives

$$P\{|X-\mu| \ge 2\} \le 0.7292,$$

where $\mu$ is the mean of $X$. Compare the bound with the exact probability and comment.

  1. Mean of a fair die. $\mu = E(X) = \dfrac{1+2+3+4+5+6}{6} = \dfrac{21}{6} = 3.5$.
  2. Second moment. $E(X^2) = \dfrac{1+4+9+16+25+36}{6} = \dfrac{91}{6} = 15.1667$.
  3. Variance. $\sigma^2 = E(X^2) - \mu^2 = 15.1667 - (3.5)^2 = 15.1667 - 12.25 = 2.9167$.
  4. Apply Chebyshev in the form $P\{|X-\mu| \ge c\} \le \dfrac{\sigma^2}{c^2}$ with $c = 2$: $$P\{|X-3.5| \ge 2\} \le \frac{2.9167}{4} = 0.7292 \qquad \blacksquare$$
  5. Exact probability. $|X-3.5| \ge 2$ means $X \le 1.5$ or $X \ge 5.5$, and since $X$ takes only whole values that is $X \in \{1, 6\}$: $$P = \frac{2}{6} = 0.3333$$
  6. Comment. $0.3333 \le 0.7292$, so the inequality holds — and it is loose by more than a factor of two. That is expected: Chebyshev is built to hold for every distribution with this mean and variance, so it must leave room for far nastier ones than a die. It guarantees a ceiling, never an estimate.
σ² = 2.9167 · Chebyshev bound = 0.7292 · exact = 0.3333 · bound holds, and is loose
Question 3 — Sampling distribution of $\bar{X}$ 2+2+2 marks
Easy Straight CLT — the guaranteed marks on this paper

According to a telecom operator, customer-service calls last on average 4.5 minutes. Assume the population standard deviation is 1.2 minutes. Suppose a random sample of 36 calls is taken.

a. What is the probability that the sample mean is more than 4.8 minutes?

b. What is the probability that the sample mean is less than 4.2 minutes?

c. What is the probability that the sample mean is between 4.3 and 4.9 minutes?

  1. Standard error once, used three times. $n = 36 \ge 30$, so by the Central Limit Theorem $\bar{X}$ is approximately normal with $$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{36}} = \frac{1.2}{6} = 0.2$$
  2. (a) $z = \dfrac{4.8-4.5}{0.2} = 1.50$, so $P(\bar{X} > 4.8) = 1 - \Phi(1.50) = 1 - 0.9332 = 0.0668$.
  3. (b) $z = \dfrac{4.2-4.5}{0.2} = -1.50$, so $P(\bar{X} < 4.2) = \Phi(-1.50) = 0.0668$. (Same answer as (a), by symmetry — 4.2 and 4.8 are equidistant from 4.5.)
  4. (c) $z_1 = \dfrac{4.3-4.5}{0.2} = -1.00$ and $z_2 = \dfrac{4.9-4.5}{0.2} = 2.00$, so $$P = \Phi(2.00) - \Phi(-1.00) = 0.9772 - 0.1587 = 0.8185$$
a. 0.0668 · b. 0.0668 · c. 0.8185
Question 4 — Confidence interval for μ 6 marks
Easy–Moderate σ unknown, small sample — the t-interval

Let $\{3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15,\ 17\}$ be the set of collected random sample values from a $N(\mu, \sigma^2)$ distribution, where both $\mu$ and $\sigma^2$ are unknown.

Find the 95% confidence interval for $\mu$.

Hint: σ is unknown, so it is t, not z — and the degrees of freedom are $n-1$.

  1. Sample mean. $\sum x = 80$ and $n = 8$, so $\bar{x} = \dfrac{80}{8} = 10$.
  2. Sample variance. The deviations are $-7,-5,-3,-1,1,3,5,7$, whose squares are $49,25,9,1,1,9,25,49$ and sum to $\sum(x-\bar{x})^2 = 168$. Therefore $$s^2 = \frac{168}{8-1} = 24 \qquad s = \sqrt{24} = 4.8990$$
  3. Choose the tool. $\sigma$ is unknown and $n = 8$ is small, so use $t$ with $df = n-1 = 7$.
  4. Critical value. For a 95% two-sided interval, $t_{0.025,\,7} = 2.365$.
  5. Standard error and margin of error. $$\frac{s}{\sqrt{n}} = \frac{4.8990}{\sqrt{8}} = 1.7321 \qquad ME = 2.365 \times 1.7321 = 4.0963$$
  6. Assemble. $10 \pm 4.0963$.
95% CI for μ = (5.9037, 14.0963)
Question 5 — Proportion test, then its Type II error 4+3 marks
(a) Moderate (b) Hard p-value method, then the probability of missing a real shortfall

A supplier claimed that at least 92% of the components which he supplied are conforming to the specifications. However, a random sample of 250 components showed only 220 were conforming to the specifications.

a. Test the claim of the supplier at 5% level of significance by using the p-value method. (4)

b. Using the same test, find the probability of a Type II error if the true conforming rate is in fact 88%, and state the power of the test. (3)

  1. (a) Hypotheses. "At least 92%" carries the equality, so it is the null. The sample falls short of the claim, so the alternative points left: $$H_0: p \ge 0.92 \qquad H_1: p < 0.92$$
  2. Sample proportion. $\hat{p} = \dfrac{220}{250} = 0.8800$.
  3. Standard error under $H_0$ — use $p_0$, never $\hat{p}$. $$\sigma_{\hat{p}} = \sqrt{\frac{p_0(1-p_0)}{n}} = \sqrt{\frac{0.92 \times 0.08}{250}} = \sqrt{0.0002944} = 0.017158$$
  4. Test statistic. $z = \dfrac{0.8800-0.9200}{0.017158} = \dfrac{-0.0400}{0.017158} = -2.3313$, read as $-2.33$.
  5. p-value. The test is left-tailed, so the p-value is the whole left tail: $P(Z < -2.33) = 0.0099$.
  6. Decide and conclude. $0.0099 < 0.05$, so reject $H_0$ at the 5% level. There is sufficient evidence that fewer than 92% of components conform; the supplier's claim is not supported.
  7. (b) Find the decision boundary in $\hat{p}$ units. Left-tailed at $\alpha = 0.05$ uses $z = -1.645$, so we reject only when $$\hat{p} < 0.92 - 1.645(0.017158) = 0.92 - 0.028225 = 0.8918$$
  8. Re-centre on the true value $p = 0.88$. The standard error changes with $p$: $$\sigma_{\hat{p}}' = \sqrt{\frac{0.88 \times 0.12}{250}} = \sqrt{0.0004224} = 0.020552$$
  9. β is the chance of landing in the non-rejection region. $$\beta = P(\hat{p} > 0.8918 \mid p = 0.88) = 1 - \Phi\!\left(\frac{0.8918-0.88}{0.020552}\right) = 1 - \Phi(0.57) = 1 - 0.7157 = 0.2843$$
  10. Power. $1 - \beta = 0.7157$. So even though the true rate really is 88%, this test would fail to detect it about 28% of the time.
a. z = −2.3313, p-value = 0.0099 < 0.05, reject H₀ · b. β = 0.2843, power = 0.7157
Question 6 — Variance test, cross-checked by an interval 4+2 marks
Hard A χ² test, then the CI that must agree with it

The lifetimes (in hours) of 9 randomly chosen bulbs from a large consignment are given below.

10109951005 10009901015 9851008992

a. Can we say that the population variance of the distribution of lifetimes, from which the above sample of 9 bulbs was drawn, is equal to 100 square hours? Test an appropriate hypothesis at 5% level. (4)

b. Construct the 95% confidence interval for $\sigma^2$ and show that it agrees with your decision in part (a). (2)

  1. (a) Hypotheses. "Is it equal to 100" is non-directional, so the test is two-tailed: $$H_0: \sigma^2 = 100 \qquad H_1: \sigma^2 \ne 100$$
  2. Sample mean. $\sum x = 9000$, so $\bar{x} = \dfrac{9000}{9} = 1000$.
  3. Sum of squared deviations. The deviations from 1000 are $10, -5, 5, 0, -10, 15, -15, 8, -8$, whose squares are $100, 25, 25, 0, 100, 225, 225, 64, 64$, giving $\sum(x-\bar{x})^2 = 828$ and $$s^2 = \frac{828}{8} = 103.5$$
  4. Test statistic. $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{828}{100} = 8.2800 \qquad df = 8$$
  5. Critical values. Two-tailed at 5% puts 0.025 in each tail, so with $df = 8$: $\chi^2_{0.975,\,8} = 2.180$ and $\chi^2_{0.025,\,8} = 17.535$.
  6. Decide. $2.180 < 8.2800 < 17.535$, so the statistic lies in the non-rejection region. Do not reject $H_0$. At the 5% level there is no evidence against the claim that the population variance equals 100 square hours.
  7. (b) Build the interval from the same two critical values. $$\left(\frac{(n-1)s^2}{\chi^2_{0.025,\,8}},\ \frac{(n-1)s^2}{\chi^2_{0.975,\,8}}\right) = \left(\frac{828}{17.535},\ \frac{828}{2.180}\right) = (47.2199,\ 379.8165)$$
  8. Show the agreement. The hypothesised value 100 lies inside $(47.2199, 379.8165)$, and a two-tailed test at 5% rejects exactly when the hypothesised value falls outside the 95% interval. Both routes therefore give the same verdict: do not reject. Note also how wide the interval is — nine observations say very little about a variance.
a. χ² = 8.2800, df = 8, non-rejection region (2.180, 17.535) → do not reject H₀ · b. 95% CI for σ² = (47.2199, 379.8165), which contains 100 ✓
Question 7 — Spearman rank correlation 7 marks
Hard Repeated values — mid-ranks and correction factors

Find the rank correlation coefficient between the Advertisement Cost and Sales from the data given below.

Advertisement Cost ('000 Rs.) 25302540 35304550 3555
Sales (Lakhs Rs.) 40453860 52485870 5065

Hint: check the Advertisement Cost row for repeats before ranking anything.

  1. Spot the ties. In Advertisement Cost, 25 occurs twice, 30 occurs twice and 35 occurs twice. Sales has no repeats. Tied values must share the average of the ranks they would have occupied.
  2. Assign mid-ranks. Sorted, the costs are $25,25,30,30,35,35,40,45,50,55$ in positions 1 to 10. The two 25s occupy positions 1 and 2, so each gets $1.5$; the two 30s occupy 3 and 4, so each gets $3.5$; the two 35s occupy 5 and 6, so each gets $5.5$. Then $40 \to 7$, $45 \to 8$, $50 \to 9$, $55 \to 10$.
  3. Build the working table.
    CostSales$R_x$ $R_y$$d$$d^2$
    25401.52−0.50.25
    30453.530.50.25
    25381.510.50.25
    406078−11
    35525.56−0.50.25
    30483.54−0.50.25
    45588711
    5070910−11
    35505.550.50.25
    556510911
    $\sum d^2$5.5
    Check: $\sum d = 0$ ✓
  4. Correction factor — one term per tied group. With $C.F. = \dfrac{m(m^2-1)}{12}$ and three groups of $m = 2$: $$\frac{2(4-1)}{12} = 0.5 \quad\text{three times}\quad \Rightarrow\quad \sum C.F. = 1.5$$
  5. Apply the corrected formula. With $n = 10$, $n(n^2-1) = 990$: $$r = 1 - \frac{6}{n(n^2-1)}\left[\sum d^2 + \sum C.F.\right] = 1 - \frac{6}{990}\left[5.5 + 1.5\right] = 1 - \frac{42}{990}$$ $$= 1 - 0.0424 = 0.9576$$
  6. Interpret. A very strong positive rank correlation — heavier advertising spend goes with higher sales almost monotonically across the ten observations.
Σd² = 5.5 · ΣC.F. = 1.5 · r = 0.9576
Question 8 — Regression and correlation 2+2+2+2 marks
Hard Two lines in, everything out — then a prediction

In a partially destroyed laboratory record of an analysis of bivariate data the following results only are legible.

Legible $\operatorname{Var}(X) = 9$, and the two regression lines are $$2X - 3Y + 12 = 0 \qquad \text{and} \qquad 4X - 5Y + 6 = 0$$

a. The mean values of $X$ and $Y$.

b. Which line is the regression of $Y$ on $X$, with justification, and the coefficient of correlation between $X$ and $Y$.

c. The standard deviation of $Y$.

d. Estimate $Y$ when $X = 24$, stating which line you must use and why.

  1. (a) Both regression lines pass through $(\bar{x}, \bar{y})$, so the means are the simultaneous solution. From the first, $2\bar{x} = 3\bar{y} - 12$. Substituting into the second: $$2(3\bar{y}-12) - 5\bar{y} + 6 = 0 \;\Rightarrow\; 6\bar{y} - 24 - 5\bar{y} + 6 = 0 \;\Rightarrow\; \bar{y} = 18$$ and then $\bar{x} = \dfrac{3(18)-12}{2} = \dfrac{42}{2} = 21$. Check in the second line: $4(21) - 5(18) + 6 = 84 - 90 + 6 = 0$ ✓
  2. (b) Decide the assignment using $r^2 = b_{yx}\,b_{xy} \le 1$. Try the first line as $Y$ on $X$: $3Y = 2X + 12$ gives $b_{yx} = \dfrac{2}{3} = 0.6667$. Then the second as $X$ on $Y$: $4X = 5Y - 6$ gives $b_{xy} = \dfrac{5}{4} = 1.25$. $$r^2 = \frac{2}{3} \times \frac{5}{4} = \frac{5}{6} = 0.8333 \le 1 \quad\text{— valid}$$
  3. Confirm the other assignment is impossible. If the first were $X$ on $Y$ then $b_{xy} = \dfrac{3}{2}$, and the second as $Y$ on $X$ gives $b_{yx} = \dfrac{4}{5}$, so $r^2 = 1.5 \times 0.8 = 1.2 > 1$ — impossible. So $2X - 3Y + 12 = 0$ is the regression of $Y$ on $X$.
  4. Correlation. $r = \pm\sqrt{0.8333} = \pm 0.9129$, and both regression coefficients are positive, so $r$ takes the positive sign: $r = 0.9129$.
  5. (c) Standard deviation of Y. $\operatorname{Var}(X) = 9$ gives $\sigma_x = 3$. From $b_{yx} = r\dfrac{\sigma_y}{\sigma_x}$: $$\frac{2}{3} = 0.9129 \times \frac{\sigma_y}{3} \;\Rightarrow\; \sigma_y = \frac{2}{0.9129} = 2.1909$$ Check with the other coefficient: $b_{xy} = r\dfrac{\sigma_x}{\sigma_y} = 0.9129 \times \dfrac{3}{2.1909} = 1.25$ ✓
  6. (d) Choose the line by what is being predicted. We are given $X$ and want $Y$, so we must use the regression of $Y$ on $X$ — the line that minimises vertical deviations. That is $2X - 3Y + 12 = 0$. Using the other line here would be a straightforward error.
  7. Substitute. $$Y = \frac{2X+12}{3} = \frac{2(24)+12}{3} = \frac{60}{3} = 20.0000$$
a. x̄ = 21, ȳ = 18 · b. 2X−3Y+12=0 is Y on X, r = 0.9129 · c. σ_y = 2.1909 · d. Ŷ = 20.0000

Paper 4 score

— / 50

The 26 easy-to-moderate marks are Q1–Q4 and Q5(a); the 24 hard marks are Q5(b) to Q8. Score those two halves separately — if the easy half is not close to full marks, that is the first thing to fix, because it is the half you are guaranteed to see again.

Permitted in the examination

Statistical Tables

Standard normal · Student's t · Chi-square

Standard normal — area to the LEFT of z

Each cell is $\Phi(z) = P(Z \le z)$. Find the row for the first decimal, then the column for the second. For a right-tail area use $1 - \Phi(z)$.

Student's t — upper-tail critical values

The column heading is the area in ONE tail. For a two-sided 95% interval use the $t_{0.025}$ column; for a two-sided 99% interval use $t_{0.005}$.

Chi-square — critical values

The heading $\chi^2_{0.050}$ means the area to the RIGHT of that value is 0.050. For a two-tailed test at 5% you need the $\chi^2_{0.975}$ and $\chi^2_{0.025}$ columns.

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