Question

Suppose the lengths of pregnancies of a certain animal are approximately normally distributed with mean ? = 240 days and ? = 18 days.

a. What is the probability that a randomly selected pregnancy lasts less than 233 days?

b. Suppose a random sample of 17 pregnancies is obtained. Describe the mean and standard deviation of the distribution of the sample mean length of pregnancies.

c. What is the probability that a random sample of 17 pregnancies has a mean pregnancy of less than 233 days?

According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1200 Americans results in 108 indicating that they are afraid to fly.

a. What is the sampling distribution of the proportion of Americans who are afraid to fly?

b. What is the point estimate of the sample (?̂)?

c. What is the probability that 108 or fewer out of 1200 Americans are afraid to fly?

d. Is this evidence that the proportion of Americans who are afraid to fly is decreasing? Why or why not?

Please show all work for the following problems. If a calculator is used, include the calculator function used as well as the values entered. (Example: normalcdf

Answer 1

Ans:

1)

mean=240 and standard deviation=18

a)

z=(233-240)/18

z=-0.389

P(z<-0.389)=0.3487

b)

sapmling distribution of sample means:

mean=240

standard deviation=18/sqrt(17)=4.366

c)

z=(233-240)/4.366

z=-1.603

P(z<-1.603)=0.0544

2)

a)Normal distribution with

mean=0.10

standard deviation=sqrt(0.10*(1-0.10)/1200)=0.00866

b)

point estimate of the sample (?̂)=108/1200=0.09

c)

z=(0.09-0.10)/0.00866

z=-1.155

P(z<=-1.155)=0.1241

d)No,there is not sufficient  evidence that the proportion of Americans who are afraid to fly is decreasing(as probability in part c is greater than 0.05)