Respuesta :
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 0.0082
1 - 0.0082 = 0.9918
0.9918 = 99.18% probability that they have a mean height greater than 63.0 inches.
