# The overhead reach distances of adult females are normally distributed with a mean of 200 cm and a standard deviation of 8 cm. a. Find the probability that an individual distance is greater than 210.00 210.00 cm. b. Find the probability that the mean for 25 randomly selected distances is greater than 198.70 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

###### Question:

200 cm and a standard deviation of

8 cm.

a. Find the probability that an individual distance is greater than

210.00

210.00 cm.

b. Find the probability that the mean for

25

randomly selected distances is greater than

198.70 cm.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

## Answers

Answer:a) The probability that an individual distance is greater than 210.00 cm is 0.1056b)The probability that the mean for 25 randomly selected distances is greater than 198.70 cm is 0.7917c)The normal distribution be used in part (b), even though the sample size does not exceed 30 since the parent population from which the sample has be obtained is Normal.Step-by-step explanation:a)We let the random variable X denote the overhead reach distances of individual adult females. This implies that X is normally distributed with a mean of 200 cm and a standard deviation of 8 cm. The probability that an individual distance is greater than 210 can be written as;Pr(X>210)We standardize the value to obtain the associated z-score;[tex]P(X>210)=P(Z>\frac{210-200}{8})=P(Z>1.25)[/tex]Using the standard normal tables, the area to the right of 1.25 is 0.1056. Therefore, the probability that an individual distance is greater than 210.00 cm is 0.1056.b)The first step will be to determine the sampling distribution of the sample mean given that the variable overhead reach distance is normally distributed with a mean of 200 cm and a standard deviation of 8 cm.In this case, the sample mean will also be normally distributed with a mean of 200 cm and a standard deviation of;[tex]\frac{sigma}{\sqrt{n} }=\frac{8}{\sqrt{25} }=1.6[/tex]The probability that the mean for 25 randomly selected distances is greater than 198.70 cm;Pr(sample mean >198.70)We standardize the value to obtain the associated z-score;[tex]=P(Z>\frac{198.70-200}{1.6})=P(Z>-0.8125)[/tex]Using the standard normal tables, the area to the right of -0.8125 is 0.7917. Therefore, the probability that the mean for 25 randomly selected distances is greater than 198.70 cm is 0.7917.c)The normal distribution be used in part (b), even though the sample size does not exceed 30 since the parent population from which the sample has be obtained is Normal.