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Chapter 4: Q97E (page 262)

Variable life insurance return rates. With a variable life insurance policy, the rate of return on the investment (i.e., the death benefit) varies from year to year. A study of these variable return rates was published in International Journal of Statistical Distributions (Vol. 1, 2015). A transformedratio of the return rates (x) for two consecutive years was shown to have a normal distribution, with 渭=1.5 and role="math" localid="1660283206727" 蟽=0.2. Use the standard normal table or statistical software to find the following probabilities.

a.P(1.3<x<1.6)

b. P(x>1.4)

c. P(x<1.5)

Short Answer

Expert verified
  1. P1.3<x<1.6=0.5328
  2. role="math" localid="1660283402755" Px>1.4=0.6915

c. Px<1.5=0.50

Step by step solution

01

Given information

Variable life insurance return rates followa normal distribution with渭=1.5 and蟽=0.2

02

Finding the z values

z=x-渭蟽=x-1.50.2

For, x = 1.6

Then,

z=x-渭蟽=1.6-1.50.2=0.5

For, x=1.3

Then,

z=x-渭蟽=1.3-1.50.2=-1

For, x=1.4

Then,

z=x-渭蟽=1.4-1.50.2=-0.5

For, x=1.5

Then,

z=x-渭蟽=1.5-1.50.2=0

03

Probability calculation when P(1.3<x<1.6)

a.

P1.3<x<1.6=Px<1.6-Px<1.3=Pz<0.5-Pz<-1=Pz<0.5-1-Pz<-1=0.6915-1-0.8413=0.6915-1+0.8413=0.5328P1.3<x<1.6=0.5328

Therefore, the value of P(1.3<x<1.6) is 0.5328.

04

Probability calculation when P(x>1.4)

Px>1.4=1-Px<1.4=1-Pz<-0.5=1-1-Pz<-0.5=0.6915Px>1.4=0.6915

Therefore, the value ofPx>1.4 is 0.6915.

05

Probability calculation when P(x<1.5)

b.

P(x<1.5)=P(z<0)=0.50P(x<1.5)=0.50

Therefore, the value of P(x<1.5) is 0.50.

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Most popular questions from this chapter

Soft-drink dispenser. The manager of a local soft-drink bottling company believes that when a new beverage dispensing machine is set to dispense 7 ounces, it in fact dispenses an amount at random anywhere between 6.5and 7.5 ounces inclusive. Suppose has a uniform probability

distribution.

a.Is the amount dispensed by the beverage machine a discreteor a continuous random variable? Explain.

b. Graph the frequency function forX , the amount of beverage the manager believes is dispensed by the new machine when it is set to dispense 7 ounces.

c. Find the mean and standard deviation for the distribution graphed in part b, and locate the mean and theinterval 渭卤2蟽on the graph.

d. Find P(x鈮�7).

e. FindP(x<6) .

f. FindP(6.5鈮�x鈮�7.25) .

g. What is the probability that each of the next six bottles filled by the new machine will contain more than7.25 ounces of beverage? Assume that the amount of beverage dispensed in one bottle is independent of the amount dispensed in another bottle.

If x is a binomial random variable, use Table I in Appendix D to find the following probabilities:

a.for n = 10, p = .4

b.for n = 15, p = .6

c.for n = 5, p = .1

d.for n = 25, p = .7

e.for n = 15, p = .9

f.for n = 20, p = .2

Estimating demand for white bread. A bakery has determined that the number of loaves of its white bread demanded daily has a normal distribution with mean 7,200 loaves and standard deviation 300 loaves. Based on cost considerations, the company has decided that its best strategy is to produce a sufficient number of loaves so that it will fully supply demand on 94% of all days.

a. How many loaves of bread should the company produce?

b. Based on the production in part a, on what percentage of days will the company be left with more than 500 loaves of unsold bread?

Normal probability plots for three data sets are shown on the next page. Which plot indicates that the data are approximately normally distributed?

USGA golf ball specifications. According to the US. Golf Association (USGA), "The diameter of the [golf] ball must not be less than 1.680 inches" (USGA, 2016). The USGA periodically checks the specifications of golf balls by randomly sampling balls from pro shops around the country. Two dozen of each kind are sampled, and if more than three do not meet requirements, that kind of ball is removed from the USGA's approved-ball list.

  1. What assumptions must be made and what information must be known in order to use the binomial probability distribution to calculate the probability that the USGA will remove a particular kind of golf ball from its approved-ball list?
  2. Suppose 10% of all balls produced by a particular manufacturer are less than 1.680 inches in diameter and assume that the number of such balls, x, in a sample of two dozen balls can be adequately characterized by a binomial probability distribution. Find the mean and standard deviation of the binomial distribution.
  3. Refer to part b. If x has a binomial distribution, then so does the number, y, of balls in the sample that meet the USGA's minimum diameter. [Note: x + y = 24] Describe the distribution of y. In particular, what are p, q and n? Also, find E(y) and the standard deviation of y.
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