13.3 The Normal distribution as a model — Practise

Work through these on your own. Check the second number in $N(\mu, \sigma^2)$ : it is the variance. Sketch the curve and shade the region before you read the table.

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

The random variable $Z \sim N(0, 1)$ . Find $P(Z < 1.5)$ .
The random variable $Z \sim N(0, 1)$ . Find $P(Z > 2)$ .
The random variable $Z \sim N(0, 1)$ . Find $P(Z < -1)$ .
The random variable $Z \sim N(0, 1)$ . Find $P(-0.5 < Z < 1.5)$ .
The random variable $X \sim N(30, 4^2)$ . Work out the standardised value $z$ for $x = 36$ .
The mass of a bag of sugar is modelled as $X \sim N(20, 5^2)$ grams. Find $P(X < 26)$ .
The mass of a bag of sugar is modelled as $X \sim N(20, 5^2)$ grams. Find $P(X > 12)$ .
The height of a mature sunflower is modelled as $X \sim N(45, 6^2)$ cm. Find the probability that a plant is taller than 52 cm.
The height of a mature sunflower is modelled as $X \sim N(45, 6^2)$ cm. Find $P(39 < X < 51)$ .
The score in a test is modelled as $X \sim N(100, 10^2)$ . Find $P(X < 120)$ .
The random variable $X \sim N(50, 6^2)$ . Find the value $a$ such that $P(X < a) = 0.90$ .
The mass of a jar is modelled as $X \sim N(500, 8^2)$ grams. Find the probability that a jar has mass greater than 512 grams.

Answers and concise solutions

1. The probability

Answer: This is the table value directly: $P(Z < 1.5) = \Phi(1.5) = 0.9332$ .

2. The probability

Answer: Use the complement: $P(Z > 2) = 1 - \Phi(2) = 1 - 0.9772 = 0.0228$ .

3. The probability

Answer: By symmetry, $P(Z < -1) = 1 - \Phi(1) = 1 - 0.8413 = 0.1587$ .

4. The probability

Answer: $P(-0.5 < Z < 1.5) = \Phi(1.5) - \Phi(-0.5) = 0.9332 - 0.3085 = 0.6247$ .

5. The value of $z$

Answer: $z = \dfrac{x - \mu}{\sigma} = \dfrac{36 - 30}{4} = 1.5$ . The value 36 lies 1.5 standard deviations above the mean.

6. The probability

Answer: Standardise: $z = \dfrac{26 - 20}{5} = 1.2$ . Then $P(X < 26) = \Phi(1.2) = 0.8849$ .

7. The probability

Answer: Standardise: $z = \dfrac{12 - 20}{5} = -1.6$ . Then $P(X > 12) = P(Z > -1.6) = \Phi(1.6) = 0.9452$ .

8. The probability

Answer: Standardise: $z = \dfrac{52 - 45}{6} = 1.1667$ . Then $P(X > 52) = 1 - \Phi(1.1667) \approx 0.1217$ . About 12.2% of plants are taller than 52 cm.

9. The probability

Answer: Standardise both ends: $z_1 = -1$ and $z_2 = 1$ . Then $P(39 < X < 51) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6827$ . This is the one-standard-deviation band.

10. The probability

Answer: Standardise: $z = \dfrac{120 - 100}{10} = 2$ . Then $P(X < 120) = \Phi(2) = 0.9772$ .

11. The value of $a$

Answer: The value of $z$ with $\Phi(z) = 0.90$ is $z = 1.2816$ . Then $a = \mu + z\sigma = 50 + 1.2816 \times 6 = 57.69$ (2 d.p.).

12. The probability

Answer: Standardise: $z = \dfrac{512 - 500}{8} = 1.5$ . Then $P(X > 512) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$ .