Statistics Problems: Sample Size, Probabilities, Normal and Binomial Calculations

Exam A — Statistical Problems

1) Sample size to estimate a proportion

Problem: Suppose you are a retailer and you want to estimate the proportion of your customers who are shoplifters. The estimated proportion is p = 0.07. Find the required sample size to estimate p to within E = 0.03 with 90% confidence.

Formula: n = z² * p(1 - p) / E²

Given: p = 0.07, E = 0.03, 90% confidence → z = 1.645

Calculation: n = (1.645)² * 0.07*(1 - 0.07) / (0.03)² = 2.706025 * 0.0651 / 0.0009 = 0.1761622 / 0.0009 ≈ 195.74

Result: Round up to the next whole person: n = 196.

2) Probability of selling more than 5 computers

Problem: The number x of sales that a company might expect per month is given by a probability distribution (not fully shown here). Based on the provided probabilities, what is the probability that the company will sell more than 5 computers per month?

Calculation shown: p = 0.1 + 0.4 + 0.1 = 0.6.

3) Expected commission for magazine subscriptions

Problem: A mail-order magazine subscription receives orders for one-, two-, and three-year subscriptions with probabilities 0.5, 0.3, and 0.2 respectively. Each subscription year generates a $5 commission per year. What is the expected commission for each order received?

Years: 1, 2, 3

Commission $: 5, 10, 15

Probabilities: 0.5, 0.3, 0.2

Expected commission = 5*0.5 + 10*0.3 + 15*0.2 = 2.5 + 3 + 3 = $8.50.

4) Decision between two projects (expected net profit)

Problem: As assistant to the general manager you are given two projects (A and B). The decision is based on expected net profit.

Project A outcomes and probabilities:

• -100 with probability 0.1 → -100 * 0.1 = -10
• 650 with probability 0.3 → 650 * 0.3 = 195
• 700 with probability 0.4 → 700 * 0.4 = 280
• 1150 with probability 0.2 → 1150 * 0.2 = 230

Expected value (A) = -10 + 195 + 280 + 230 = 695.

Project B outcomes and probabilities:

• 400 with probability 0.2 → 400 * 0.2 = 80
• 300 with probability 0.2 → 300 * 0.2 = 60
• 600 with probability 0.3 → 600 * 0.3 = 180
• 1000 with probability 0.3 → 1000 * 0.3 = 300

Expected value (B) = 80 + 60 + 180 + 300 = 620.

Decision: Choose the project with the larger expected net profit. Project A: 695 vs Project B: 620 → select Project A.

6) Z-scores and probabilities

Problem: Compute z-scores and associated normal probabilities for given values.

Case 1: z = (0.14 - 0.20) / 0.04 = -1.50.

P(Z ≤ -1.50) ≈ 0.0668; P(Z ≥ -1.50) = 1 - 0.0668 = 0.9332.

Case 2: z = (0.21 - 0.20) / 0.04 = 0.25.

P(Z ≤ 0.25) ≈ 0.5987; P(Z ≥ 0.25) = 1 - 0.5987 = 0.4013.

Problems 7, 8, 9 — Binomial distribution (n = 15, p = 0.7)

Context: The probability that a company executive uses a computer in his job is 0.7. Fifteen company executives are surveyed independently about their computer usage behavior.

7) Probability no more than 12 use a computer

P(X ≤ 12) = 0.8732 (given).

8) Probability at least 11 use a computer

P(X ≥ 11) = 1 - P(X ≤ 10) = 1 - 0.4845 = 0.5155 (given).

9) Probability between 6 and 10 inclusive

P(6 ≤ X ≤ 10) = P(X ≤ 10) - P(X ≤ 5) (or as provided: P(X ≤ 10) - P(X ≤ 6) depending on table usage)

Using the provided values: P(X ≤ 10) = 0.4848 and P(X ≤ 6) = 0.0152, so

P(6 ≤ X ≤ 10) ≈ 0.4848 - 0.0152 = 0.4696.

11) Area under the normal curve between z = 1.25 and z = 2.25

P(1.25 < Z < 2.25) = Φ(2.25) - Φ(1.25) ≈ 0.9878 - 0.8944 = 0.0934.

12) Find z for P(Z ≥ z) = 0.1093

P(Z ≥ z) = 0.1093 ⇒ P(Z ≤ z) = 1 - 0.1093 = 0.8907.

z ≈ Φ^-1(0.8907) ≈ 1.23.

13) Probability between z = -1.53 and z = 2.20

P(-1.53 < Z < 2.20) = Φ(2.20) - Φ(-1.53) = 0.9861 - 0.0630 = 0.9231.

15 & 16) Spark plug life (normal distribution)

Given: Population mean μ = 60,000 miles, population standard deviation σ = 5,000 miles. Sample size n = 100 spark plugs.

15) Probability a spark plug lasts more than 60,900 miles

Compute z = (60,900 - 60,000) / 5,000 = 900 / 5,000 = 0.18.

P(X > 60,900) = 1 - Φ(0.18) ≈ 1 - 0.5714 = 0.4286.

16) 95% confidence interval for the mean life

CI: μ ± z*(σ / √n) = 60,000 ± 1.96 * (5,000 / 10) = 60,000 ± 1.96 * 500 = 60,000 ± 980.

95% CI: [59,020, 60,980].

Exam B — Additional Problems

1) Sample size for p = 0.07, E = 0.02 at 95% confidence

Given: p = 0.07, E = 0.02, 95% confidence ⇒ z = 1.96.

n = (1.96)² * 0.07*(1 - 0.07) / (0.02)² = 3.8416 * 0.0651 / 0.0004 = 0.25008816 / 0.0004 ≈ 625.22

Result: Round up to whole person: n = 626.

4) Probability fewer than 4 computers

Provided calculation: P = 0.19 + 0.15 + 0.08 + 0.02 = 0.62.

3) Expected commission with $6 per subscription-year

Probabilities: 0.5 (1 year), 0.3 (2 years), 0.2 (3 years).

Commissions: $6, $12, $18

Expected commission = 6*0.5 + 12*0.3 + 18*0.2 = 3 + 3.6 + 3.6 = $10.20.