Let [math]X[/math] and [math]Y[/math] be the number of hours that a randomly selected person watches movies and sporting events, respectively, during a three-month period. The following information is known about [math]X[/math] and [math]Y[/math]:

• [math]\operatorname{E}(X) = 50 [/math]
• [math]\operatorname{E}(Y) = 20 [/math]
• [math]\operatorname{Var}(X) = 50 [/math]
• [math]\operatorname{Var}(Y) = 30 [/math]
• [math]\operatorname{Cov}(X,Y) = 10 [/math]

The totals of hours that different individuals watch movies and sporting events during the three months are mutually independent.

One hundred people are randomly selected and observed for these three months. Let [math]T[/math] be the total number of hours that these one hundred people watch movies or sporting events during this three-month period.

Approximate the value of [math]\operatorname{P}[T \lt 7100][/math].

• 0.62
• 0.84
• 0.87
• 0.92
• 0.97

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: B

Observe that (where [math]Z[/math] is total hours for a randomly selected person)

[[math]] \operatorname{E}[ Z ] = \operatorname{E}[ X + Y ] = \operatorname{E}[ X ] + \operatorname{E}[Y] = 50 + 20 = 70, [[/math]]

[[math]] \operatorname{Var}[ Z ] = \operatorname{Var}[ X + Y ] = \operatorname{Var}[ X ] + \operatorname{Var}[Y ] + 2\operatorname{Cov}[ X , Y ] = 50 + 30 + 20 = 100. [[/math]]

It then follows from the Central Limit Theorem that [math]T[/math] is approximately normal with mean 100(70) = 7000 and variance 100(100) = 10,000 and standard deviation 100. The probability of being less than 7100 is the probability that a standard normal variable is less than (7100 –7000)/100 = 1. From the tables, this is 0.8413.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.