Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 247 days and standard deviation sigma equals 16 days. Complete parts (a) through (f) below.

The answer is given below

Step-by-step explanation:

a) What is the probability that a randomly selected pregnancy lasts less than 242 days

First we have to calculate the z score. The z score is used to determine the measure of standard deviation by which the raw score is above or below the mean. It is given by:

Given that Mean (μ) = 247 and standard deviation (σ) = 16 days. For x < 242 days,

From the normal distribution table, P(x < 242) = P(z < -0.3125) = 0.3783

(b) Suppose a random sample of 17 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies.

If a sample of 17 pregnancies is obtained, the new mean the new standard deviation:

c) What is the probability that a random sample of 17 pregnancies has a mean gestation period of 242 days or less

From the normal distribution table, P(x < 242) = P(z < -1.29) = 0.0985

d) What is the probability that a random sample of 49 pregnancies has a mean gestation period of 242 days or less?

From the normal distribution table, P(x < 242) = P(z < -2.19) = 0.0143

(e) What might you conclude if a random sample of 49 pregnancies resulted in a mean gestation period of 242 days or less?

It would be unusual if it came from mean of 247 days

f) What is the probability a random sample of size 2020 will have a mean gestation period within 11 days of the mean

For x = 236 days

For x = 258 days

From the normal distribution table, P(236 < x < 258) = P(-3.07 < z < 3.07) = P(z < 3.07) - P(z < -3.07) =0.9985 - 0.0011 = 0.9939

5.25 ounces

step-by-step explanation:

divided 3/4 and 6