Using the normal distribution, it is found that there is a 0.1357 = 13.57% probability that the total amount paid for these second movies will exceed $15.00.
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.
- For n instances of a normal variable, the mean is [tex]n\mu[/tex] and the standard error is [tex]s = \sigma\sqrt{n}[/tex]
In this problem:
- Mean of $0.47, standard deviation $0.15, hence [tex]\mu = 0.47, \sigma = 0.15[/tex]
- 30 instances, hence [tex]n\mu = 30(0.47) = 14.1, s = 0.15\sqrt{30} = 0.8216[/tex]
The probability is 1 subtracted by the p-value of Z when X = 15, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Considering the n instances:
[tex]Z = \frac{X - n\mu}{s}[/tex]
[tex]Z = \frac{15 - 14.1}{0.8216}[/tex]
[tex]Z = 1.1[/tex]
[tex]Z = 1.1[/tex] has a p-value of 0.8643.
1 - 0.8643 = 0.1357.
0.1357 = 13.57% probability that the total amount paid for these second movies will exceed $15.00.
A similar problem is given at https://brainly.com/question/25769446
