Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean = 125 days and standard deviation 12 days, Complete parts (a) through ( below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (b) Suppose a random sample of 20 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies The sampling distribution of is with s-ando:-D (Round to four decimal places as needed) (c) What is the probability that a random sample of 20 pregnancies has a mean gestation period of 121 days or low? The probability that the mean of a random sample of 20 pregnancies is less than 121 days is approximately (Round to four decimal places as needed) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) O A. If 100 independent random samples of size n = 20 pregnancies were obtained from this population, we would expect 121 days or less B. If 100 independent random samples of size n= 20 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of exactly 121 days C. If 100 independent random samples of size n = 20 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of 121 days or more (d) What is the probability that a random sample of 51 pregnancies has a mean gestation period of 121 days or less? sample(s) to have a sample mean of a Click to select your answer(s).

Answer 1

(b)

By Central limit theorem,

the sampling distribution of
T
is normal with $\mu_{\bar{x}}$ = 125 and $\sigma_{\bar{x}}$ = 12 / $\sqrt{20}$ = 2.6833

(c)

P( < 121) = P[Z < (121 - 125)/2.6833]

T

= P[Z < -1.49]

= 0.0681

For 100 samples, expected proportion = 100 * 0.0681 = 6.81 $\approx$ 7

A. If 100 independent random samples of size n = 20 pregnancies were obtained from this population, we would expect 7 samples to have a sample mean of 121 or less.

(d)

The sampling distribution of
T
is normal with $\mu_{\bar{x}}$ = 125 and $\sigma_{\bar{x}}$ = 12 / $\sqrt{51}$ = 1.680336

P( < 121) = P[Z < (121 - 125)/1.680336]

T

= P[Z < -2.38]

= 0.0087

For 100 samples, expected proportion = 100 * 0.0087  = 0.87 $\approx$ 1

C. If 100 independent random samples of size n = 51 pregnancies were obtained from this population, we would expect 1 samples to have a sample mean of 121 or less.

(e)

The result would be unusual so the sample likely came from a population whose mean gestation period is less than 125 days

(f)

The sampling distribution of
T
is normal with $\mu_{\bar{x}}$ = 125 and $\sigma_{\bar{x}}$ = 12 / $\sqrt{15}$ = 3.098387

P(125 - 8 < < 125+8) = P[-8/3.098387 < Z < 8/3.098387]

T

= P[2.58 < Z < -2.58] = P[Z < 2.58] - P[Z < -2.58]

= 0.9951 - 0.0049

= 0.9902