Question

provide an appropriate response. use the standard normal table to find the probability. assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. the cheerleaders for a local professional basketball team must be between 65.5 and 68.0 inches. if a woman is randomly selected, what is the probability that her height is between 65.5 and 68.0 inches?

a. 0.7881
b. 0.1844
c. 0.9608
d. 0.3112

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 63.6$ (mean), $\sigma = 2.5$ (standard deviation).
For $x = 65.5$, $z_1=\frac{65.5 - 63.6}{2.5}=\frac{1.9}{2.5}=0.76$.
For $x = 68.0$, $z_2=\frac{68.0 - 63.6}{2.5}=\frac{4.4}{2.5}=1.76$.

Step2: Use the standard - normal table

We want to find $P(0.76From the standard - normal table, $P(Z < 1.76)=0.9608$ and $P(Z < 0.76)=0.7764$.
Then $P(0.76

Answer:

B. 0.1844