Question

Heights of men on a baseball team have a​ bell-shaped distribution with a mean of 178 cm178 cm and a standard deviation of 5 cm5 cm. Using the empirical​ rule, what is the approximate percentage of the men between the following​ values?
a. 168168 cm and 188188 cm
b. 163163 cm and 193193 cm

Answer 1

Answer on Question #57529 - Math – Statistics and Probability

Question

Heights of men on a baseball team have a bell-shaped distribution with a mean of 178 cm and a standard deviation of 5 cm. Using the empirical rule, what is the approximate percentage of the men between the following values?

a. 168 cm and 188 cm

b. 163 cm and 193 cm

Solution

If the mean of the standard normal distribution is $m = 178$ and the standard deviation is $\sigma = 5$, then

$$\begin{array}{l} P(m - \sigma \leq \xi \leq m + \sigma) = P(178 - 5 \leq \xi \leq 178 + 5) = P(173 \leq \xi \leq 183) = 0.6827, \\ P(m - 2\sigma \leq \xi \leq m + 2\sigma) = P(178 - 2 \cdot 5 \leq \xi \leq 178 + 2 \cdot 5) = P(168 \leq \xi \leq 188) = \\ = 0.9545, \end{array}$$ $$\begin{array}{l} P(m - 3\sigma \leq \xi \leq m + 3\sigma) = P(178 - 3 \cdot 5 \leq \xi \leq 178 + 3 \cdot 5) = P(163 \leq \xi \leq 193) = \\ = 0.9973 \text{ according to the empirical rule}. \end{array}$$

a. The percentage of men in range between 168 cm and 188 cm is given by

$$\frac{1}{5\sqrt{2\pi}} \int_{168}^{188} e^{-\frac{(x - 178)^2}{2 \cdot 5^2}} dx = \operatorname{erf}(\sqrt{2}) = 0.9545$$

Thus, the percentage is 95.45%.

b. The percentage of men in range between 168 cm and 188 cm is given by

$$\frac{1}{5\sqrt{2\pi}} \int_{163}^{193} e^{-\frac{(x - 178)^2}{2 \cdot 5^2}} dx = 0.9973$$

Thus, the percentage is 99.73%.

Answer: a. 95.45%, b. 99.73%.

www.AssignmentExpert.com