1. A recent survey found that 55% of pet owners admit to talking to their pet on

Question

1. A recent survey found that 55% of pet owners admit to talking to their pet on the phone or leaving messages for the pet on the answering machine. If 300 pet owners are randomly selected, what is the probability that 150 or more will admit to talking to their pet on the phone or leaving messages?

2. Management of an airline uses a normal distribution to model the value claimed for a lost piece of luggage on domestic flights. The mean of the distribution $600 and the standard deviation is $85. Suppose a random sample of 64 pieces of luggage is to be selected. What is the probability that the average claim value for the 64 pieces is less than $580.

3.Some managers of companies use employee rankings to evaluate their workers. Suppose the distribution of rankings of employees at large company is normal with a mean of 65 points and a standard deviation of 6 points.
A) What percent of workers score below 50%?

B)The managers of the company need to cut back on the number of employees, so they plan to fire those who score in the lowest 10%. What is the lowest score an employee could have and keep their job?

C) Suppose the company has 20 employees. What is the probability that the average ranking is at least 70 points?

D) Suppose it is known that only 3% of all employees earn bonuses by having a ranking score above 77. Suppose this particular company has 195 employees. What is the probability that more than 4%of their employees will earn a bonus?

Explanation / Answer

1. Given p=0.55, n=300
mean=n*p=300*0.55 =165
standard deviation =n*p*(1-p)=sqrt(300*0.55*(1-0.55)) = 8.62

So the probability is

P(X>150) = P((x-mean)/s > (150-165)/8.62)

=P(Z> -1.74)

=0.9591 (check standard normal table)

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2. Given X~Normal(mean=600, s=85)

n=64

So the probability is

P(xbar<580) = P((xbar-mean)/(s/n) < (580-600)/(85/8))

=P(Z<-1.88)

= 0.0301 (check standard normal table)

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3. Given X~Normal(mean=65, s=6)

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A) What percent of workers score below 50%?

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B)The managers of the company need to cut back on the number of employees, so they plan to fire those who score in the lowest 10%. What is the lowest score an employee could have and keep their job?

P(X<c)=0.1

--> P((X-mean)/s <(c-65)/6) =0.1

--> P(Z<(c-65)/6)=0.1

--> (c-65)/6 = -1.282 (check standard normal table)

--> c= 65 -1.282*6 =57.308

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C) Suppose the company has 20 employees. What is the probability that the average ranking is at least 70 points?

P(xbar>70) = P((xbar-mean)/(s/n) > (70-65)/(6/sqrt(20)))

=P(Z>3.73)

=0.00009

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