Statistics

Managerial Statistics: Assignment II

1. An insurance company sells a $20,000 whole life insurance policy for an annual premium of $300. Actuarial tables show that a person who would be sold such a policy with this premium has a 0.001 probability of death during a year. Let X be an random variable (RV) representing the insurance company’s profit made on one of these policies during a year. Find the expected profit and standard deviation for the insurance company. The probability distribution of X is:

X, Profit | P(X) |

$300 (if policyholder lives) | 0.999 |

$300-$20,000 = -$19,700(if policyholder dies) | 0.001 |

Answer:

The expected value ie profit (X) can be calculated as μx = $ 300 (0.999) + (- $ 19,700)(0.001) = $ 280

Variance is 0.249001. So, Std deviation is 0.499.

2. The following questions (a through d) are based on the table that represents the daily demand for pies (X) at Christina’s Pie Shop in Brooklyn, NY. Assume each day is independent from others.

X Probability

5 0.117

6 0.235

7 0.256

8 0.185

9 0.108

10 0.099

a. What is the average number of pies sold per day?

b. What is the standard deviation of the number of pies sold per day?

c. What is the mode number of pies sold per day?

d. What is the probability that Christina sells at least 7 pies in any given day?

Answer:

a) Average = 5(0.117)+6(0.235)+7(0.256)+8(0.185)+9(0.108)+10(0.099)= 7.229

b) Std deviation is 1.469

c) Mode is 7 because it has the most probability ie 0.256

d) Probability to sell > 7 every day is P(X > 7) = 0.256+0.185+0.108+0.099 = 0.648

3. Find the area under a standard normal curve between the mean and the point z = - 1.26. What does this area represent?

Answer:

The area between the mean and z of – 1.26 is 39.6 %. Represents the probability a random variable will be between the mean and standard deviation of...