Normal distribution and annuities

1) A company that sells annuities must base the annual payout on the probability distribution of the length of life of the participants in the plan. Suppose the probability distribution of the lifetimes of the participants is approximately a normal distribution with a mean of 68 years and a standard deviation of 3.5 years. What proportion of the plan recipients would recipients would receive payment beyond age 75
A) 2.28% (B) 09772 (C) 09997 (D) 0.8413
P(X > 75) = 1 - NORMDIST(75,68,3.5,1) = 0.02275 = 2.28%

2) Suppose Z has a standard normal distribution with a mean of 0 and a standard deviation of 1.0 One-half 50% of the possible Z values are between____ and ____ symmetrically distributed about the mean.
A) -0.50 and + 0.50 (B) -0.69 and + 0.69 (C) 0-67 and +0.67 (D) -1.96 and +1.96

LCL = NORMSINV(0.025) = -1.96
UCL = -NORMSINV(0.025) = 1.96

3) The probability that a standard normal random variable, Z is between 1.00 and 3.00 is
A) 0.1574 (B) 0.3158 (C) 1.33 (D) 2.0
P(1 < Z < 3) =NORMSDIST(3)-NORMSDIST(1)= 0.157305

4) The following ordered array(from left to right) depicts the amount of money (in dollars) withdrawn from a cash machine by 25 customers at a local bank:

40 50 50 70 70
80 80 90 100 100
100 100 100 100 110
110 120 120 130 140
140 150 160 160 200

Compute descriptive summary measure from data and determine which of the following sentences best describes the distributions
A) The distribution is perfectly symmetrical (B) distribution is slightly right-skewed (C) distribution is slightly left-skewed (D) Distribution is highly asymmetrical

5) A state lottery is conducted in which six winning numbers are selected from a total of 54 numbers. What is the probably that, if six numbers are randomly selected, all six numbers will be winning numbers
A) 0.000038719 (B) 0.000011111 (C) 0.000000039 (D) 0.000000004
P = BINOMDIST(6,6,6/54,0) = 0.00001

6) Two investments, which we will call X and Y, have the following characteristics:
Expected to return of X=$50 Expected return of Y=$100
Variance of X=9,000Variance of Y =15,000
Covariance of X and Y=7,500

If the weight assigned to investment X of portfolio assets is 0.4, what is the portfolio risk
A) 40% (B) 314.643 (C) 80 (D) 56.92

7) Using the company records for the past 500 working days, the manager of Koning Motors, a suburban automobile dealership , has summarized the number of cars sold per day into the following tables
# Cars Sold Frequency Product
0 40 0
1 100 100
2 142 284
3 66 198
4 36 144
5 30 150
6 26 156
7 20 140
8 16 128
9 14 126
10 8 80
11 2 22
Total 500 1528
Mean = 1528/500 = 3.056

What is the mean or expected number of cars sold per day?
A) 3.056 (B) 3.136(C) 4.545(D) 4.069

8) For some positive value of Z, the probability that a standard normal variable is between 0 and Z is 0.3770. The value of Z is:
A) 0.31 (B) 0.81 (C) 1.16 (D) 1.47
A = -NORMSINV(0.377) = 0.31

9) The number of power outages at a nuclear power plant has a Poisson distribution with a variance of six outages per year. What is the probability that there will be no more than one outage in a year?
A) 0.0149 (B) 0.0174 (C) 0.1667 (D) Cannot be solved because the mean (expected number of successes) is not given

10) In its standardized form, the normal distribution
A) has a mean of 0 and a standard deviation of 1 (B) has a mean of 1 and a variance of 0 (C) has an area equal 0.5 (D) cannot be used to approximate discrete probability distribution

11) Thirty-six of the staff of 80 teachers at a local intermediate school are certified in Cardio-Pulmonary Resuscitation (CPR). In 180 days of school, assuming that none of the teachers were ever absent, about how many days is it that the teacher on the bus duty will likely be certified in CPR?
A) 5 days (B) 45days (C) 65days (D) 81days
Proportion = 36/80 = 0.45
If there are 180days, the required proportion = 0.45*180 = 81 days

12) What type of probability distribution will the consulting firm most likely employ to analyze the insurance claims in the following problem?

An insurance company has called a consulting firm to determine if a company has an unusually high number of false insurance claims. It is known that the industry proportion for false claims is 3%. The consulting firm has decided to randomly and independently sample 100 of the company's insurance claims. They believe the number of these that are false will yield the information the company desires.

A) Normal Distribution (B) Binomial distribution (C) Poisson Distribution (D) hypergeometric distribution