Chapter 2.2, Problem 59E

(a)

To determine

To find: the deciles of the standard Normal distribution

Answer to Problem 59E

The decilesof the standard Normal distribution are $\pm 1.28$ .

Explanation of Solution

Given:

The deciles of distribution are lowest 10% and the highest 10%

The deciles of a density curve are with area 0.1 and 0.9 to their left

Calculation:

  

The deciles of the standard Normal Curve is the value of z with 10% and 90% of the distribution to its left. The total area of the standard Normal Curve is 1 and since curve is symmetrical, the area from $z=0$ to either end is 0.5. Here, 0.1 can be written as $0.5-0.4$ and 0.9 can be written as $0.5+0.4$ , Using Table of Normal Curve, the entry closest to value 0.4 is 0.3997. This is the entry corresponding to $z=1.28$ . So $z=-1.28$ is the standardized value with area 0.10 to its left and $z=1.28$ is the standardized value with area 0.90 to its left. Hence, the deciles are $\pm 1.28$ .

Conclusion:

Hence, the deciles are $\pm 1.28$ .

(b)

To determine

To find: the deciles of the distribution

Answer to Problem 59E

The deciles for the heights of young women are 61.3 inches and 67.7 inches.

Explanation of Solution

Given:

The deciles of distribution are lowest 10% and the highest 10%

The deciles of a density curve are with area 0.1 and 0.9 to their left

Meanis 64.5 inches

Standard deviation is 2.5 inches.

Calculation:

Suppose the random variable x denotes heights of young women, then it is given that:

  $x~N\left(\mu ,{\sigma }^2\right)$ , where $\mu =64.5$ inches and $\sigma =2.5$ inches

The standardized value of unknown xis $z=\pm 1.28$ . So x satisfies the equation:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \pm 1.28=\frac{x-64.5}{2.5}\end{array}$

Solving for x gives:

  $x=61.3\text{inches}$ or $\text{x=67}\text{.7inches}$

The deciles for the heights of young women are 61.3 inches and 67.7 inches.

Conclusion:

Therefore, the deciles for the heights of young women are 61.3 inches and 67.7 inches.