MATH SOLVE

4 months ago

Q:
# Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 197 days and standard deviation sigma equals 16 days. Complete parts (a) through (c). (a) What is the probability that a randomly selected pregnancy lasts less than 192 days? The probability that a randomly selected pregnancy lasts less than 192 days is approximately . 3783. (Round to four decimal places as needed.) (b) What is the probability that a random sample of 29 pregnancies has a mean gestation period of less than 192 days? The probability that the mean of a random sample of 29 pregnancies is less than 192 days is approximately nothing

Accepted Solution

A:

Answer:a) 0.3783 is the probability that a random sample of 29 pregnancies has a mean gestation period of less than 192 days.b) 0.046 is the probability that the mean of a random sample of 29 pregnancies is less than 192 daysStep-by-step explanation:We are given the following information in the question:
Mean, μ = 197Standard Deviation, σ = 16We are given that the distribution of lengths of the pregnancies is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(pregnancy lasts less than 192 days)
P(x < 192)
[tex]P( x < 192) = P( z < \displaystyle\frac{192 - 197}{16}) = P(z < =-0.3125)[/tex]
Calculation the value from standard normal z table, we have, [tex]P(x < 192) = 0.3783 = 37.83\%[/tex]
b) P( a random sample of 29 pregnancies has a mean period of less than 192 days)
P(x < 192)
[tex]P( x < 192) = P( z < \displaystyle\frac{192 - 197}{\frac{16}{\sqrt{29}}}) = P(z < = -1.6828)[/tex]
Calculation the value from standard normal z table, we have, [tex]P(x < 192) = 0.046 = 4.6\%[/tex]