Practice Problems with Solutions

• Step 1 — Mean. Sum = 12+15+18+22+15+20+25+18+15+20 = 180; n = 10; x̄ = 180/10 = 18.0

• Step 2 — Median. Sorted: 12, 15, 15, 15, 18, 18, 20, 20, 22, 25. n is even, so average the 5th and 6th: (18 + 18)/2 = 18.0

• Step 3 — Mode. 15 appears three times, more than any other value. Mode = 15

• Step 4 — Range. 25 − 12 = 13

• Step 5 — Variance.

s² = 136 / (10 − 1) = 136/9 = 15.11

• Step 6 — Standard deviation. s = √15.11 = 3.888

• Step 7 — Coefficient of variation. CV = (3.888/18.0) × 100 = 21.6%

Note that mean = median = 18 while the mode is 15 — the distribution is close to symmetric, with a slight cluster at the lower end.

A multiple-choice test has 12 questions, each with 4 options. A student guesses every answer. Find the probability of (a) exactly 5 correct, (b) at most 2 correct, and (c) the mean and variance of the number correct.

Solution. Binomial with n = 12, p = 1/4 = 0.25.

• (a) P(X = 5) = ¹²C₅ × 0.25⁵ × 0.75⁷
• ¹²C₅ = 792
• 0.25⁵ = 0.0009766
• 0.75⁷ = 0.1334839

• P = 792 × 0.0009766 × 0.1334839 = 0.1032

• (b) P(X ≤ 2) = P(0) + P(1) + P(2)

• P(0) = 0.75¹² = 0.031676
• P(1) = 12 × 0.25 × 0.75¹¹ = 0.126705
• P(2) = 66 × 0.0625 × 0.75¹⁰ = 0.232292

• P(X ≤ 2) = 0.3907

• (c) Mean = np = 12 × 0.25 = 3.0; Variance = np(1−p) = 12 × 0.25 × 0.75 = 2.25

So a guesser expects 3 right out of 12, and has about a 39% chance of getting 2 or fewer.

• (a) z = (1400 − 1200)/150 = 200/150 = 1.3333 P(Z > 1.3333) = 1 − 0.9088 = 0.0912, about 9.1%

• (b) z₁ = (1000 − 1200)/150 = −1.3333; z₂ = (1300 − 1200)/150 = 0.6667 P = 0.7475 − 0.0912 = 0.6563, about 65.6%

• (c) The 10th percentile has z = −1.2816 x = μ + zσ = 1200 + (−1.2816)(150) = 1200 − 192.2 = 1007.8 hours

So 10% of bulbs fail before about 1008 hours.

• Step 1. x̄ = 30/5 = 6; ȳ = 160/5 = 32

• Step 2.

• Step 3. r = 166 / √(40 × 698) = 166 / √27920 = 166 / 167.09 = 0.9935

Interpretation: a very strong positive linear relationship — as advertising spend rises, sales rise almost proportionally. This does not prove that advertising causes the sales; both could be driven by seasonal demand.

(a) The regression line.

• b₁ = Σ(x−x̄)(y−ȳ) / Σ(x−x̄)² = 166/40 = 4.15
• b₀ = ȳ − b₁x̄ = 32 − 4.15(6) = 32 − 24.9 = 7.1
• ŷ = 7.1 + 4.15x

Interpretation: each additional ₹1 lakh of advertising is associated with about ₹4.15 lakh more sales. The intercept of 7.1 suggests ₹7.1 lakh of sales with no advertising — but x = 0 is outside the observed range of 2 to 10, so treat it cautiously.

(b) Prediction at x = 7.

ŷ = 7.1 + 4.15(7) = 7.1 + 29.05 = ₹36.15 lakh

x = 7 lies inside the observed range, so this is interpolation and is reasonably safe.

(c) Residuals and R².

• SS_total = Σ(y−ȳ)² = 698.00
• SS_residual = 9.10
• SS_regression = 698.00 − 9.10 = 688.90
• R² = 688.90/698.00 = 0.9870

So 98.70% of the variation in sales is explained by advertising spend.

Check: r² = 0.9935² = 0.9870 ✓

(d) ANOVA table.

• df_regression = k = 1; df_residual = n − 2 = 3; df_total = n − 1 = 4
• F = 688.90 / 3.0333 = 227.11

(e) Testing the slope.

• H₀: β₁ = 0 (advertising has no linear effect on sales)
• H₁: β₁ ≠ 0

• α = 0.05, df = 3

• SE(b₁) = √(MS_res / Σ(x−x̄)²) = √(3.0333/40) = √0.075833 = 0.27538

• t = b₁ / SE(b₁) = 4.15 / 0.27538 = 15.0703
• Critical value t(0.025, 3) = 3.182
• p-value = 0.000634

Since |15.07| > 3.182 and p < 0.05, reject H₀.

Check: t² = 15.0703² = 227.11 = F ✓

Conclusion. There is very strong evidence of a linear relationship between advertising spend and sales. The model explains 98.7% of the variation, and the slope is highly significant.

Step 1 — Hypotheses.

• H₀: μ = 1000 (the company's claim is correct)
• H₁: μ ≠ 1000 (the mean lifetime differs from the claim)

Two-tailed, because the question asks whether the claim is wrong, not specifically whether bulbs last less.

Step 2 — Significance level. α = 0.05.

Step 3 — Choose the test. The population standard deviation is unknown and n = 16 is small, so use the one-sample t-test with df = 16 − 1 = 15.

Step 4 — Test statistic.

• standard error = s/√n = 80/√16 = 80/4 = 20
• t = (x̄ − μ₀)/SE = (960 − 1000)/20 = −40/20 = −2.00

Step 5 — Decision.

• Critical values: t(0.025, 15) = ±2.131
• p-value = 0.0639

|−2.00| = 2.00 < 2.131, and p = 0.0639 > 0.05 → fail to reject H₀.

Step 6 — Conclusion. At the 5% significance level there is not enough evidence to reject the company's claim. The observed shortfall of 40 hours could plausibly arise from sampling variation in a sample of only 16 bulbs.

Two things worth adding, which earn marks:

• Note how close this is. At α = 0.10 the critical value is 1.753 and we would reject. The conclusion depends on a threshold chosen in advance — which is exactly why α must be fixed before seeing the data.

• "Fail to reject H₀" is not "the claim is true". A larger sample might well detect a real shortfall. With n = 16 the test simply lacks the power.

Step 1 — Hypotheses.

• H₀: language preference is independent of year of study
• H₁: they are associated

Step 2 — α = 0.05.

Step 3 — Expected frequencies. E = (row total × column total)/grand total

All expected frequencies are ≥ 5 ✓

Step 4 — Test statistic.

Every observed value equals its expected value exactly, so every term (O − E)²/E is zero:

χ² = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 = 0.00

Step 5 — Decision.

• df = (3 − 1)(3 − 1) = 4
• Critical value χ²(0.05, 4) = 9.488
• p-value = 1.000

0.00 < 9.488 → fail to reject H₀.

Step 6 — Conclusion. There is no evidence whatsoever of an association. Language preference is independent of year of study.

Why this problem is instructive: the proportions are identical in every row — each year splits 50% Python, 25% Java, 25% C++. Perfect independence gives χ² = 0 exactly. Real data never does this, so if you compute χ² = 0 on an exam question, check whether the data really is proportionally identical (as here) or whether you have made an arithmetic slip.