13.3 The Normal distribution as a model — Practise with help

Work each question closed book first. Sketch the curve and shade the region before you read the table. Check that your probability lies between 0 and 1.

Warm-up: remember

The random variable $X \sim B(8, 0.5)$ . Find $P(X \le 1)$ .
A data set has mean 40 and standard deviation 5. A value is 47. How many standard deviations above the mean is it?
The random variable $X$ has mean 20 and standard deviation 4. Standardise the value $x = 26$ .

Answers

$P(X \le 1) = \dfrac{1 + 8}{256} = \dfrac{9}{256} \approx 0.0352$ .
$\dfrac{47 - 40}{5} = 1.4$ , so the value is 1.4 standard deviations above the mean.
$z = \dfrac{26 - 20}{4} = 1.5$ .

If question 1 was hard, revise the binomial from Section 13.1.

Practise with help

1. The random variable $Z \sim N(0, 1)$ . Find $P(Z < 1.2)$ .

Hint

This is the value the table gives directly. Read the row for $z = 1.2$ .

Next step

Find $z = 1.2$ in the standard Normal table and read $\Phi(1.2)$ .

Full working

From the table,

P(Z < 1.2) = \Phi(1.2) = 0.8849.

So about 88.5% of the area lies to the left of 1.2.

2. The random variable $Z \sim N(0, 1)$ . Find $P(Z > 1.5)$ .

Hint

The table gives the area to the left. The area to the right is the complement.

Next step

Use $P(Z > 1.5) = 1 - \Phi(1.5)$ , and read $\Phi(1.5)$ from the table.

Full working

P(Z > 1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668.

So about 6.7% of the area lies to the right of 1.5.

3. The random variable $Z \sim N(0, 1)$ . Find $P(-1 < Z < 2)$ .

Hint

Subtract the area to the left of $-1$ from the area to the left of 2.

Next step

Use $\Phi(-1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$ . Then subtract.

Full working

P(-1 < Z < 2) = \Phi(2) - \Phi(-1) = 0.9772 - 0.1587 = 0.8185.

So about 81.9% of the area lies between $-1$ and 2.

4. The random variable $X \sim N(30, 4^2)$ . Find $P(X < 36)$ .

Hint

Standardise $x = 36$ first. The standard deviation is 4.

Next step

$z = \dfrac{36 - 30}{4} = 1.5$ . Now read $\Phi(1.5)$ from the table.

Full working

Standardise:

z = \frac{36 - 30}{4} = 1.5.

Then

P(X < 36) = P(Z < 1.5) = \Phi(1.5) = 0.9332.

5. The random variable $X \sim N(200, 10^2)$ . Find $P(X > 215)$ .

Hint

Standardise $x = 215$ , then use the complement.

Next step

$z = \dfrac{215 - 200}{10} = 1.5$ . The table gives the area to the left, so use $P(X > 215) = 1 - \Phi(1.5)$ .

Full working

Standardise:

z = \frac{215 - 200}{10} = 1.5.

Then

P(X > 215) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668.

6. The random variable $X \sim N(50, 6^2)$ . Find the value $a$ such that $P(X < a) = 0.95$ .

Hint

Start from the probability and find $z$ first. The table of common probabilities gives $z = 1.6449$ for $p = 0.95$ .

Next step

Unstandardise with $a = \mu + z\sigma = 50 + 1.6449 \times 6$ .

Full working

The value of $z$ with $\Phi(z) = 0.95$ is $z = 1.6449$ . Then

a = \mu + z\sigma = 50 + 1.6449 \times 6 = 50 + 9.8694 = 59.87 \ (2 \text{ d.p.}).

So 95% of the values lie below 59.87.