Question

The heights of 1000 students are approximately normally distributed with a mean of 154.5cm and a standard deviation of 3.0cm.
a. If a student is selected at random, what is the probability that his height is more than 158cm?
b. How many of the students will be more than 158cm tall?
c. Obtain the probability that a randomly selected student has a height between 150cm and 160cm.
d. What is the probability that a random sample of size 36 will have a mean height of more than 155cm?

Answer 1

Answer on Question #82838 – Math – Statistics and Probability

The heights of 1000 students are approximately normally distributed with a mean of 154.5cm and a standard deviation of 3.0cm.

Question

a. If a student is selected at random, what is the probability that his height is more than 158cm?

Solution

$$\begin{array}{l} X \sim N(154.5, 3.0^2) \\ z = \frac{x - \mu}{\sigma} \\ z = \frac{158 - 154.5}{3} = 1.167 \\ P(X > 158) = 1 - P(X \leq 158) = 1 - P(Z \leq 1.167) = 1 - 0.8783 = 0.1217 \\ \end{array}$$

Question

b. How many of the students will be more than 158cm tall?

Solution

$$\begin{array}{l} P(X > 158) = 0.1217 \\ N = 1000 \\ 0.1217 \cdot 1000 \approx 122 \text{ students} \\ \end{array}$$

Question

c. Obtain the probability that a randomly selected student has a height between 150cm and 160cm.

Solution

$$\begin{array}{l} X \sim N(154.5, 3.0^2) \\ z = \frac{x - \mu}{\sigma} \\ z_1 = \frac{150 - 154.5}{3} = -1.500 \\ z_2 = \frac{160 - 154.5}{3} = 1.833 \\ P(150 < X < 160) = P(X < 160) - P(X \leq 150) = \\ \end{array}$$ $$= P(Z < 1.833) - P(Z \leq -1.500) \approx 0.9666 - 0.0668 = 0.8998$$

Question

d. What is the probability that a random sample of size 36 will have a mean height of more than $155\mathrm{cm}$?

Solution

$$\mu_{\overline{X}} = \mu_X = 154.5$$ $$\sigma_{\overline{X}} = \frac{\sigma_X}{\sqrt{n}} = \frac{3.0}{\sqrt{36}} = 0.5$$ $$z = \frac{\overline{x} - \mu_{\overline{X}}}{\sigma_{\overline{X}}} = \frac{155 - 154.5}{0.5} = 1.000$$ $$P(\overline{X} > 155) = 1 - P(\overline{X} \leq 155) = 1 - P(Z \leq 1.000) = 1 - 0.8413 = 0.1587.$$

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