Question

Question Help Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean u 243 days and standard deviation a 15 days. Complete parts (a) through () below. (a) What is the probability that a randomly selected pregnancy lasts less than 237 days? The probability that a randomly selected pregnancy lasts less than 237 days is approximately(Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fillin the answer box within your choice (Round to the nearest integer as needed.) O A. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last exactly 237 days. OB. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last less than 237 days. O C. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last more than 237 days. (b) Suppose a random sample of 17 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. with ;and The sampling distribution of x is (Round to four decimal places as needed.) (e) What is the probability that a random sample of 17 pregnancies has a mean gestation period of 237 days or less? The probability that the mean of a random sample of 17 pregnancies is less than 237 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.) A. If 100 independent random samples of size n 17 pregnancies were obtained from this population, we would expect of exactly 237 days. sample(s) to have a sample mean O B. of 237 days or more. If 100 independent random samples of size n 17 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean |sample(s) to have a sample mean Oc. If 100 independent random samples of size n 17 pregnancies were obtained from this population, we would expect of 237 days or less (d) What is the probability that a random sample of 29 pregnancies has a mean gestation period of 237 days or less? The probability that the mean of a random sample of 29 pregnancies is less than 237 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 29 pregnancies were obtained from this population, we would expect of 237 days or less sample(s) to have a sample mean O B. If 100 independent random samples of size n 29 pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of 237 days or more. O C. If 100 independent random samples of size n 29 pregnancies were obtained from this population, we would expect of exactly 237 days. sample(s) to have a sample mean (e) What might you conclude if a random sample of 29 pregnancies resulted in a mean gestation period of 237 days or less? so the sample likely came from a population whose mean gestation period is 243 days This result would be () What is the probability a random sample of size 18 will have a mean gestation period within 11 days of the mean? The probability that a random sample of size 18 will have a mean gestation period within 11 days of the mean is (Round to four decimal places as needed.)

Answer 1

Solution: Given that μ = 243, σ = 15

(a) P(X < 237) = P((X-μ)/σ < (237-243)/15)
= P(Z < -0.4)
= 0.3446

=> option B. we would expect 34.46 pregnancies to last less than 237 days

(b) The sampling distribution of x is normai with μx =243 and σx = 3.6380
=> μx = μ = 243, σx = σ/sqrt(n) = 15/sqrt(17) = 3.6380

(c) For n = 17, P(X < 237) = P((X-μ)/(σ/sqrt(n)) < (237-243)/(15/sqrt(17))
= P(Z < -1.6492)
= 0.0495

=> option C. we would expect 4.95 samples

(d) For n = 29, P(X < 237) = P((X-μ)/(σ/sqrt(n)) < (237-243)/(15/sqrt(29))
= P(Z < -2.1541)
= 0.0158

=> option A. option A. we would expect 1.58 samples

(d) Ths result wound be unusual so the sample likely came from a population whose mean gestation period is less 243 days.

(f) for n = 18, with in 11

=> P(-11/(15/sqrt(18)) < Z < (11/(15/sqrt(18))
= P(-3.1113 < Z < 3.1113)
= 0.9982