Answer :
Answer:
93.18% probability that they have a mean height between 62.9 inches and 64.0 inches.
Step-by-step explanation:
To solve this question, we need 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], the sample means with size n of at least 30 can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 63.6, \sigma = 2.5, n = 90, s = \frac{2.5}{\sqrt{90}} = 0.2635[/tex]
If 90 women are randomly selected, find the probability that they have a mean height between 62.9 inches and 64.0 inches.
This is the pvalue of Z when X = 64 subtracted by the pvalue of Z when X = 62.9. So
X = 64
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{64 - 63.6}{0.2635}[/tex]
[tex]Z = 1.52[/tex]
[tex]Z = 1.52[/tex] has a pvalue of 0.9357
X = 62.9
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{62.9 - 63.6}{0.2635}[/tex]
[tex]Z = -2.66[/tex]
[tex]Z = -2.66[/tex] has a pvalue of 0.0039
0.9357 - 0.0039 = 0.9318
93.18% probability that they have a mean height between 62.9 inches and 64.0 inches.