13.5 Selecting an appropriate distribution — Practise

Work through these on your own. For each question, state the model and give one reason before you calculate. Decide whether the quantity is a count or a measurement.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A fair coin is tossed 20 times. The random variable $X$ is the number of heads. State which model fits, and find $P(X = 10)$ .
The mass of an apple is modelled as $X \sim N(120, 15^2)$ grams. Find $P(X > 140)$ .
A machine produces faulty items with probability 0.04. A sample of 25 items is checked. The random variable $X$ is the number of faulty items. State which model fits, and find $P(X = 0)$ .
The time taken by a runner to finish a race is modelled as $X \sim N(60, 10^2)$ minutes. Find $P(X < 70)$ .
A bag holds 6 red counters and 4 blue counters. Three counters are taken without replacement. The random variable $X$ is the number of red counters. Explain why neither the binomial nor the Normal model fits, and find $P(X = 0)$ .
A basketball player scores a free throw with probability 0.9. She takes 10 free throws. The random variable $X$ is the number she scores. State which model fits, and find $P(X \ge 9)$ .
A measurement is modelled as $X \sim N(30, 5^2)$ . Find $P(25 < X < 35)$ .
A fair die is rolled 6 times. The random variable $X$ is the number of sixes. State which model fits, and find $P(X \ge 1)$ .
The mass of a jar is modelled as $X \sim N(500, 20^2)$ grams. Find $P(X > 540)$ .
The mass of a fish is modelled as $X \sim N(2.4, 0.3^2)$ kilograms. Find $P(2.1 < X < 2.7)$ .
A quiz has 10 questions, each answered true or false. A student guesses every answer. The random variable $X$ is the number correct. State which model fits, and find $P(X \le 2)$ .
A machine produces faulty items with probability 0.03. A sample of 40 items is checked. The random variable $X$ is the number of faulty items. State which model fits, and find $P(X = 0)$ .

Answers and concise solutions

1. The model and the probability

Answer: The quantity is a count in a fixed 20 independent trials with constant $p = 0.5$ , so $X \sim B(20, 0.5)$ . Then $P(X = 10) = \dbinom{20}{10}(0.5)^{20} = \tfrac{46189}{262144} \approx 0.1762$ .

2. The probability

Answer: The mass is a measurement, so the Normal model fits. Standardise: $z = \dfrac{140 - 120}{15} \approx 1.333$ . Then $P(X > 140) = 1 - \Phi(1.333) \approx 0.0912$ .

3. The model and the probability

Answer: The quantity is a count in a fixed sample of 25 with constant probability 0.04, so $X \sim B(25, 0.04)$ . Then $P(X = 0) = (0.96)^{25} \approx 0.3604$ .

4. The probability

Answer: The time is a measurement, so the Normal model fits. Standardise: $z = \dfrac{70 - 60}{10} = 1$ . Then $P(X < 70) = \Phi(1) = 0.8413$ .

5. The probability

Answer: The probability of red changes after each draw, so $p$ is not constant and the binomial model fails. $X$ is a whole-number count from 0 to 3, so the Normal model fails. Every counter is blue, and there are $\binom{4}{3} = 4$ ways to choose three blues from four against $\binom{10}{3} = 120$ selections in total, so $P(X = 0) = \tfrac{4}{120} = \tfrac{1}{30} \approx 0.0333$ .

6. The model and the probability

Answer: The quantity is a count in a fixed 10 independent trials with constant $p = 0.9$ , so $X \sim B(10, 0.9)$ . Then $P(X \ge 9) = P(9) + P(10) \approx 0.7361$ .

7. The probability

Answer: Standardise both ends: $z = -1$ and $z = 1$ . Then $P(25 < X < 35) = \Phi(1) - \Phi(-1) = 0.6827$ . The interval is the one-standard-deviation band.

8. The model and the probability

Answer: The quantity is a count in a fixed 6 independent trials with constant $p = \tfrac16$ , so $X \sim B(6, \tfrac16)$ . Use the complement: $P(X \ge 1) = 1 - \left(\tfrac56\right)^6 = \tfrac{31031}{46656} \approx 0.6651$ .

9. The probability

Answer: Standardise: $z = \dfrac{540 - 500}{20} = 2$ . Then $P(X > 540) = 1 - \Phi(2) = 1 - 0.9772 = 0.0228$ .

10. The probability

Answer: Standardise both ends: $z = \dfrac{2.1 - 2.4}{0.3} = -1$ and $z = \dfrac{2.7 - 2.4}{0.3} = 1$ . Then $P(2.1 < X < 2.7) = \Phi(1) - \Phi(-1) = 0.6827$ .

11. The model and the probability

Answer: The quantity is a count in a fixed 10 independent trials with constant $p = 0.5$ , so $X \sim B(10, 0.5)$ . Then $P(X \le 2) = \tfrac{1 + 10 + 45}{1024} = \tfrac{56}{1024} = \tfrac{7}{128} \approx 0.0547$ .

12. The model and the probability

Answer: The quantity is a count in a fixed sample of 40 with constant probability 0.03, so $X \sim B(40, 0.03)$ . Then $P(X = 0) = (0.97)^{40} \approx 0.2957$ .