1- Business Weekly conducted a survey of graduates from 30 top MBA programs. On

Question

1-

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 125000 dollars. Assume the standard deviation is 42000 dollars. Suppose you take a simple random sample of 80 graduates.

Find the probability that a single randomly selected policy has a mean value between 118426 and 125939.1 dollars.
P(118426 < X < 125939.1) = (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of size n=80n=80 has a mean value between 118426 and 125939.1 dollars.
P(118426 < M < 125939.1) = (Enter your answers as numbers accurate to 4 decimal places.)

2- I have the answers but I could not figure out how get the same results?

CNNBC recently reported that the mean annual cost of auto insurance is 984 dollars. Assume the standard deviation is 286 dollars. You take a simple random sample of 51 auto insurance policies.

Find the probability that a single randomly selected value is less than 969 dollars.
P(X < 969) = 0.4791

Find the probability that a sample of size n=51n=51 is randomly selected with a mean less than 969 dollars.
P(M < 969) = 0.354

Enter your answers as numbers accurate to 4 decimal places.

3-

Let X represent the full height of a certain species of tree. Assume that X has a normal probability distribution with mean 97.6 ft and standard deviation 80.4 ft. You intend to measure a random sample of n = 242 trees. The bell curve below represents the distribution of these sample means. The scale on the horizontal axis is the standard error (standard deviation) of the sampling distribution. Complete the indicated boxes, correct to two decimal places.

Explanation / Answer

Ans:

1)

a)

z(118426 )=(118426-125000)/42000=-0.16

z(125939.1 )=(125939.1-125000)/42000=0.02

P(-0.16<=z<=0.02)=P(z<=0.02)-P(z<=-0.16)

=0.5089-0.4378=0.0711

b)standard error of mean=42000/sqrt(80)=4695.743

z(118426 )=(118426-125000)/4695.743=-1.4

z(125939.1 )=(125939.1-125000)/4695.743=0.2

P(-1.4<=z<=0.2)=P(z<=0.2)-P(z<=-1.4)

=0.5793-0.0808=0.4985

2)

a)z=(969-984)/286=-0.05245

P(z<-0.05245)=0.4791

b)standard error of mean=286/sqrt(51)=40.05

z=(969-984)/40.05=-0.3746

P(z<-0.3746)=0.354

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