provide an appropriate response. use the standard normal table to find the probability. assume that the…

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?\n\na. 0.7881\nb. 0.1844\nc. 0.9608\nd. 0.3112

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?\n\na. 0.7881\nb. 0.1844\nc. 0.9608\nd. 0.3112

Answer

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 $P(0.76<Z<1.76)$. From the standard - normal table, $P(Z < 1.76)=0.9608$ and $P(Z < 0.76)=0.7764$. Then $P(0.76<Z<1.76)=P(Z < 1.76)-P(Z < 0.76)=0.9608 - 0.7764=0.1844$.

Answer:

B. 0.1844