Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean u = 282 days and standard deviation o = 20 days. Complete parts (a) through (f) below. (a) What is the probability that a randomly selected pregnancy lasts less than 274 days? The probability that a randomly selected pregnancy lasts less than 274 days is approximately 0.3446. (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 pregnant individuals were selected independently from this population, we would expect pregnancies to last more than 274 days. B. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last exactly 274 days. If 100 pregnant individuals were selected independently from this population, we would expect 34 pregnancies to last less than 274 days. (b) Suppose a random sample of 18 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. with The sampling distribution of x is normal (Round to four decimal places as needed.) Mã = 282 and os = 4.7140 (c) What is the probability that a random sample of 18 pregnancies has a mean gestation period of 274 days or less? The probability that the mean of a random sample of 18 pregnancies is less than 274 days is approximately (Round to four decimal places as needed.)

Answer 1

Parts a) and b) are correct

Part c) We define the standard random variable Z as $Z={\bar{X}-282 \over 4.714}$ Where denotes mean.

X

Using normal table we have

$\Pr(\bar{X} <274)=\Pr\left ( Z< {274-282 \over 4.714} \right )=\Pr(Z<-1.697) =0.0448$

So required probability is 0.0448