13.5 Selecting an appropriate distribution — Practise with help

Work each question closed book first. Decide whether the quantity is a count or a measurement, and write one sentence giving the reason for your model.

Warm-up: remember

The random variable $Z \sim N(0, 1)$ . Find $P(Z < 0.5)$ .
The random variable $X \sim B(6, 0.5)$ . Find $P(X = 2)$ .
A data set has mean 30 and standard deviation 4. A value is 38. How many standard deviations above the mean is it?

Answers

From the table, $\Phi(0.5) = 0.6915$ .
$P(X = 2) = \dbinom{6}{2}\left(\tfrac12\right)^6 = \tfrac{15}{64} \approx 0.2344$ .
$\dfrac{38 - 30}{4} = 2$ standard deviations.

If question 3 was hard, revise the standard deviation from Section 11.5.

Practise with help

1. A fair coin is tossed 20 times. The random variable $X$ is the number of heads. State which model fits the count, and find $P(X = 10)$ .

Hint

Check the four binomial conditions. Is the quantity a count or a measurement?

Next step

There are 20 independent trials, two outcomes, and $p = 0.5$ is constant. So $X \sim B(20, 0.5)$ . Use the binomial formula with $r = 10$ .

Full working

The quantity is a count of heads in a fixed 20 tosses, so the binomial model fits:

X \sim B(20, 0.5).

P(X = 10) = \binom{20}{10}(0.5)^{10}(0.5)^{10} = \binom{20}{10}(0.5)^{20} = \frac{46189}{262144} \approx 0.1762.

2. The mass of an apple from an orchard is modelled as $X \sim N(120, 15^2)$ grams. Find $P(X > 140)$ .

Hint

The mass is a measurement, so the Normal model fits. Standardise $x = 140$ .

Next step

$z = \dfrac{140 - 120}{15} \approx 1.333$ . Use the table or a calculator, then subtract from 1.

Full working

z = \frac{140 - 120}{15} = 1.333 \ (3 \text{ d.p.}).

P(X > 140) = 1 - \Phi(1.333) \approx 1 - 0.9088 = 0.0912.

So about 9.1% of the apples have mass greater than 140 grams.

3. 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 $P(X = 0)$ .

Hint

The number of trials is fixed at 25, and there are two outcomes per item. Use the binomial model with $p = 0.04$ .

Next step

$X \sim B(25, 0.04)$ , so $P(X = 0) = (0.96)^{25}$ . Work out the power and round to four decimal places.

Full working

The quantity is a count in a fixed sample of 25, so

X \sim B(25, 0.04).

P(X = 0) = (0.96)^{25} \approx 0.3604.

So about 36% of samples of 25 contain no faulty item.

4. The time taken by a runner to finish a race is modelled as $X \sim N(60, 10^2)$ minutes. Find $P(X < 70)$ .

Hint

The time is a measurement, so the Normal model fits. Standardise $x = 70$ .

Next step

$z = \dfrac{70 - 60}{10} = 1$ , so read $\Phi(1)$ from the table.

Full working

z = \frac{70 - 60}{10} = 1.

P(X < 70) = \Phi(1) = 0.8413.

So about 84% of runners finish in under 70 minutes.

5. 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 $P(X = 0)$ .

Hint

For the binomial model, look at how the probability of red changes between draws. For the Normal model, remember that $X$ is a count from 0 to 3.

Next step

The probability of red changes after each draw, so the trials are not independent and the binomial model fails. $X$ is a small whole-number count, so the Normal model fails. Count the selections with no red counter.

Full working

The probability of a red counter changes after each draw, so $p$ is not constant. The binomial model needs a constant $p$ , so it does not fit. $X$ takes the whole-number values 0, 1, 2 and 3, so a Normal curve does not fit either.

To find $P(X = 0)$ , every counter must be blue. There are $\binom{4}{3} = 4$ ways to choose three blue counters from the four, and $\binom{10}{3} = 120$ ways to choose any three counters. So

P(X = 0) = \frac{4}{120} = \frac{1}{30} \approx 0.0333.

6. A quiz has 10 questions, each answered true or false. A student guesses every answer. State which model fits, and find $P(X \le 2)$ , where $X$ is the number correct.

Hint

Each guess is a trial with two outcomes and constant probability 0.5. The quantity is a count.

Next step

$X \sim B(10, 0.5)$ , so add the terms for 0, 1 and 2. Every term has denominator $2^{10} = 1024$ .

Full working

The quantity is a count in a fixed 10 trials, so $X \sim B(10, 0.5)$ . Then

P(X \le 2) = \frac{\binom{10}{0} + \binom{10}{1} + \binom{10}{2}}{1024} = \frac{1 + 10 + 45}{1024} = \frac{56}{1024} = \frac{7}{128} \approx 0.0547.