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 a 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?
a. 0.91

b. 0.09

c. 0.14

d. 0.86

Answer 1

Answer:

c. 0.14

Step-by-step explanation:

Generally, for a given normal distribution, we can estimate the probability of the normal distribution using the equation below:

P (x greater than $x_{min}$) = pnorm($x_{min}$, mean, sd, lower tail=FALSE)

The following variables are provided in the question:

$x_{min}$ = 15

n = 30

Therefore:

Mean: E(30*X) = 30(0.47)

Standard deviation: 0.15*$\sqrt{30}$

P (y greater than 15) = 1 - pnorm (15, 30*0.47, 0.15*$\sqrt{30}$)

P (y greater than 15) = 1 - pnorm (15, 14.1, 0.822) = 0.137

The approximate probability is 0.14