The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8.6 cm.

Question

a. Find the probability that an individual distance is greater than 210.90 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.20 cm.196.20 cm.

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

Answer 1

I am not sure if you're accustomed to using a z-score and table of probabilities or technology, so I will answer using both!

a) Since you're choosing one person from the population, you will use the mean and standard deviation from the population. z= (210.9-197.5)/8.6 = 1.56 Using a table of normal probabilities,

P(x>210.9)= 1-P(z<1.56)= 1- 0.9406=0.0594

Using technology: normalcdf(210.9, 1E99, 197.5, 8.6)=0.0596

b) Since you're calculating the population for a sample to occur, you need to get a mean of the sampling distribution and standard deviation of the sampling distribution.

Mean_pop=Mean_{sampling dist.}=197.5

St. Dev_{sampling dist.}= Std_population/√n = 8.6/√15 = 2.22

Using technology, P(X>196.20)= normalcdf( 196.20, 1E99, 197.5, 2.22) = 0.7209

c) You only have to worry about the sample size in a sampling distribution if you either do not know the shape of the distribution of the population or if the distribution of the population is not normally distributed. Since this population is normally distributed (as stated in the given information), then your sample size does not have to be greater than 30.

Hope this helps! I can solve part b using a z-score if you need me to.