Answer :
Answer:
0.2206 = 22.06% probability that the sample mean is greater than 60.
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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 58, \sigma = 13, n = 25, s = \frac{13}{\sqrt{25}} = 2.6[/tex]
Compute the probability the sample mean is greater than 60:
This probability is 1 subtracted by the pvalue of Z when X = 60. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{60 - 58}{2.6}[/tex]
[tex]Z = 0.77[/tex]
[tex]Z = 0.77[/tex] has a pvalue of 0.7794
1 - 0.7794 = 0.2206
0.2206 = 22.06% probability that the sample mean is greater than 60.