Question

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 0.15*sqrt of 30 and not 0.15*30 because I thought this is a linear transformation

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• 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 $n\mu$ and the standard error is $s = \sigma\sqrt{n}$

In this problem:

• Mean of $0.47, standard deviation $0.15, hence $\mu = 0.47, \sigma = 0.15$

• 30 instances, hence $n\mu = 30(0.47) = 14.1, s = 0.15\sqrt{30} = 0.8216$

The probability is 1 subtracted by the p-value of Z when X = 15, hence:

$Z = \frac{X - \mu}{\sigma}$

Considering the n instances:

$Z = \frac{X - n\mu}{s}$

$Z = \frac{15 - 14.1}{0.8216}$

$Z = 1.1$

$Z = 1.1$ 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 brainly.com/question/25769446