Respuesta :
Answer:
13.04% probability that the sample mean of sales per customer is between $76 and $77 dollars.
Step-by-step explanation:
To solve this problem, 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, 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 = 75, \sigma = 6, n = 39, s = \frac{6}{\sqrt{39}} = 0.96[/tex]
What is the probability that the sample mean of sales per customer is between $76 and $77 dollars?
This is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 76. So
X = 77
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{77 - 75}{0.96}[/tex]
[tex]Z = 2.08[/tex]
[tex]Z = 2.08[/tex] has a pvalue of 0.9812
X = 76
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{76 - 75}{0.96}[/tex]
[tex]Z = 1.04[/tex]
[tex]Z = 1.04[/tex] has a pvalue of 0.8508
0.9812 - 0.8508 = 0.1304
13.04% probability that the sample mean of sales per customer is between $76 and $77 dollars.
