Answer :
Using the normal distribution and the central limit theorem, it is found that there is a:
a) 0.2266 = 22.66% probability that the sample mean is greater than 63.
b) 0.1587 = 15.87% probability that the sample mean is less than 56.
c) 0.6147 = 61.47% probability that the sample mean is between 56 and 63.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 60, thus [tex]\mu = 60[/tex].
- Standard deviation of 12, thus [tex]\sigma = 12[/tex].
- Sample size of 9, thus [tex]n = 9, s = \frac{12}{\sqrt{9}} = 4[/tex].
Item a:
This probability is 1 subtracted by the p-value 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 - 60}{4}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a p-value of 0.7734.
1 - 0.7734 = 0.2266
0.2266 = 22.66% probability that the sample mean is greater than 63.
Item b:
This probability is the p-value of Z when X = 56, so:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{56 - 60}{4}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.1587.
0.1587 = 15.87% probability that the sample mean is less than 56.
Item c:
This probability is the p-value of Z when X = 63 subtracted by the p-value of Z when X = 56.
These p-values were found in the previous items, so:
0.7734 - 0.1587 = 0.6147
0.6147 = 61.47% probability that the sample mean is between 56 and 63.
A similar problem is given at https://brainly.com/question/24663213