Nicole Becker – Assignment 3

Complete the following textbook exercises:

10, 18, 24, 32, 36, 54, 60 and 74

10. The mean of a normal probability distribution is 60; the standard deviation is 5.

a) What percent of the observations lie between 55 and 65?

b) What percent of the observations lie between 50 and 70?

c) What percent of the observations lie between 45 and 75?

  1. P(55<X<65) = P( -1<z <1) = .6826
  2. P(50<X<70) = P( -2<z<2) = .9545
  3. P( 45<x<75) = P(-3<z <3) = .9973

18. A normal population has a mean of 12.2 and a standard deviation of 2.5.

a) Compute the z-value associated with 14.3

b) What percent of the population is between 12.2 and 14.3?

c) What percent of the population is less than 10.0?

  1. Z= ( 14.3-12.2)/2.5= 0.84
  2. P( 12.2<X< 14.3) = P( 0<x< 0.84) = 0.299
  3. P(x <10) = P( z < -.88) = .18943

24. The amounts of money requested on home loan applications at Down River Federal Savings follow the normal distribution, with a mean of $70,000 and a standard deviation of $20,000. A loan application is received this morning. What is the probability:

a) The amount requested is $80,000 or more?

b) The amount requested is between $65,000 and $80,000?

c) The amount requested is $65,000 or more?

  1. P(X >80000)= P( z > 0.5)= 0.691462
  2. P(65000<X<80000) = P(-.25 < z < .5) = 0.290169
  3. P(x > 65000)= P(z > -.25)= 1-0.401294 = 0.598706

32. The monthly sales of mufflers follow the normal distribution with a mean of 1200 and a standard deviation of 225. The manufacturer would like to establish inventory levels such that there is only a 5 percent chance of running out of stock. Where should the manufacturer set the inventory levels?

We want x1 such that P(x > x1) = .05

As per normal tables x1= 1.645

So 1.645 = (x1 – 1200)/225

X1=1570.125

36. Assume a binomial probability distribution with n = 40 and p = .55. Compute the following:

a) The mean and standard deviation of the random variable.

b) The probability that X is 25 or more.

c) The probability that X is 15 or less.

d) The probability that X is between 15 and 25 inclusive.

N= 40

P=.55

  1. Mean= np =40*.55 =22

Since mean > 5 we can use a normal approximation for this binomial with mena = 22 and std dev=√ 40*.55*.45 = 3.146

  1. P(x >25) = P( z > (25-22)/3.146) ) = P(z>.9536)= 0.170143
  1. P(x<15) = P( z < (15-22)/3.146) ) = P(z<-2.225)= 0.013041
  1. P(15 <x <25) = P(-2.225< z<0.9536) = P(z>.9536)= 0.816816

54. The annual commissions earned by sales representatives of Machine Products Inc., a manufacturer of light machinery, follow the normal distribution. The mean yearly amount earned is $40,000 and the standard deviation is $5000.

a) What percent of the sales representatives earn more than $42,000 per year?

b) What percent of the sales representatives earn between $32,000 and $42,000?

c) What percent of the sales representatives earn between $32,000 and $35,000?

d) The sales manager wants to award the sales representatives who earn the largest commissions a bonus of $1000. He can award a bonus to 20% of the representatives. What is the cutoff point between those who earn a bonus and those who do not?

  1. P(x>42000) = P( z > 0.4)=0.344578
  2. P( 32000<x <42000)= P(-1.6<z < 0.4)=0.600622
  3. P(32000< x < 35000) = P( -1.6<z<-.333)=0.103856
  4. We want x1 such that P(x> x1) = 0.2 = P( z> z1)

Z1= .842

x1= (5000*.842) +40000 = 442010

60. A recent study shows that 20% of all employees prefer their vacation time during March break. If a company employs 50 people, what is the probability that:

a) Fewer than 5 employees want their vacation during March break?

b) More than 5 employees want their vacation during March break?

c) Exactly 5employees want their vacation during March break?

d) More than 5 but fewer than 15 employees want their vacation during March break?

This is a binomial distribution with n=50 p=.2

  1. P( x<5)= 0.01849
  2. P(x >5) = .95197
  3. P( x= 5) = .02953
  4. P(5<x<15) .01143

74. In establishing warranties on HDTV sets, the manufacturer wants to set the limits so that few will need repair at manufacturer expense. On the other hand, the warranty period must be long enough to make the purchase attractive to the buyer. For a new HDTV the mean number of months until repairs are needed is 36.84 with a standard deviation of 3.34 months. Where should the warranty limits be set so that only 10 percent of large screen TV’s need repairs at the manufacturer’s expense?

We want x1 such that P(x>x1) =.9=P(z>z1)

z1= 1.28 so that x1= (1.28*3.34)+36.84 = 41.1152