Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu = 283 days and standard deviation sigma = 29 days. (a) What is the probability that a randomly selected pregnancy lasts less than 273 days? The probability that a randomly selected pregnancy lasts less than 273 days is approximately (Round to four decimal places as needed.) (b) What is the probability that a random sample of 11 pregnancies has a mean gestation period of 273 days or less? The probability that the mean of a random sample of 11 pregnancies is less than 273 days is approximately (c) What is the probability that a random sample of 42 pregnancies has a mean gestation period of 273 days or less? The probability that the mean of a random sample of 42 pregnancies is less than 273 days is approximately (Round to four decimal places as needed.)

Answer 1

X: Lengths of the pregnancies

Mean = = 283

Standard deviation = = 29

a) We have to find P(X < 273)

For finding this probability we have to find z score.

That is we have to find P(Z < - 0.34)

P(Z < - 0.34) = 0.3651 ( Using z table)

The probability that a randomly selected pregnancy lasts less than 273 days is 0.3651

b) Sample size = n = 11

We have to find P( < 273)

For finding this probability we have to find z score.

That is we have to find P(Z < - 1.14)

P(Z < - 1.14) = 0.1264 ( Using z table)

The probability that the mean of a random sample of 11 pregnancies is less than 273 days is approximately 0.1264

c) Sample size = n = 42

We have to find P( < 273)

For finding this probability we have to find z score.

That is we have to find P(Z < - 2.23)

P(Z < - 2.23) = 0.0127 ( Using z table)

The probability that the mean of a random sample of 42 pregnancies is less than 273 days is approximately 0.0127