Assume that the heights of adult caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. if 100 women are randomly​ selected, find the probability that they have a mean height greater than 63.0 inches. round to four decimal places.

Answer:0.9918 = 99.18% probability that they have a mean height greater than 63.0 inches.Step-by-step explanation:To solve this question, we have to understand the normal probability distribution and the central limit theorem.Normal probability distribution:Problems of normally distributed samples are solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.Central limit theorem:The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation, which is also called standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]In this problem, we have that:[tex]\mu = 63.6, \sigma = 2.5, n = 100, s = \frac{2.5}{\sqrt{100}} = 0.25[/tex]Find the probability that they have a mean height greater than 63.0 inches.This is 1 subtracted by the pvalue of Z when X = 63. So[tex]Z = \frac{X - \mu}{\sigma}[/tex]By the Central Limit Theorem[tex]Z = \frac{X - \mu}{s}[/tex][tex]Z = \frac{63 - 63.6}{0.25}[/tex][tex]Z = -2.4[/tex][tex]Z = -2.4[/tex] has a pvalue of 0.00821 - 0.0082 = 0.99180.9918 = 99.18% probability that they have a mean height greater than 63.0 inches.