Every Thursday, Matt and Dave's Video Venture has “roll-the-dice" day. A customer may choose to roll two fair dice and rent a second movie for an
amount (in cents) equal to the numbers uppermost on the dice, with the larger number first. For example, if the customer rolls a two and a four, a
second movie may be rented for $0.42. If a two and two are rolled, a second movie may be rented for $0.22. Let X represent the amount paid for a
second movie on roll-the-dice day. The expected value of X is $0.47 and the standard deviation of X is $0.15.
If a customer rolls the dice and rents a second movie every Thursday for 30 consecutive weeks, what is the approximate probability that the total
amount paid for these second movies will exceed $15.00?

I saw other questions for this but why is the standard deviation 0.15sqrt of 30 and not 0.1530 because I thought this is a linear transformation

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