Question

28) The weight of adult koalas living in a particular region is normally distributed with a mean of 21.1 pounds and a standard deviation of 3.24 pounds. A sample of 6 adult koalas from the region is selected at random.

Find the probability that the mean weight of koalas in the sample is within 0.6 pounds of the population mean weight.

Round your answer to 4 decimal places.

Please explain all steps to solution in detail. Thank you!

Answer 1

Given,

= 21.1 , = 3.24

Using central limit theorem,

P( < x) = P(Z < (x - ) / ( / sqrt(n) ) )

We have to calculate P( - 0.6 < ​​​​​​ < + 0.6) = ?

= P(21.1 - 0.6 < 21.1 + 0.6) = ?

P( 20.5 < < 21.7 ) = P( < 21.7) - P(  < 20.5)

= P(Z < (21.7 - 21.1) / (3.24 / sqrt(6) )) - P(Z < (20.5 - 21.1) / (3.24 / sqrt(6) ))

= P(Z < 0.45) - P(Z < -0.45)

= 0.6736 - 0.3264 (From Z table)

= 0.3472