Question

In a population of Siberian flying squirrels in western Finland, assume that the the number of pups born to each female over her lifetime has mean μ=3.66μ=3.66 and standard deviation σ=2.9598.σ=2.9598. The distribution of squirrel pups born is non‑normal because it takes only whole, non‑negative values.

Determine the mean number of pups, ¯¯¯¯¯X,X¯, such that in 90%90% of all random samples of such squirrels of size n=60,n=60, the mean number of pups born to females in the sample is less than ¯¯¯¯¯X.than X¯.

You may find the table of ZZ‑critical values useful.

Give your answer to at least two decimal places.

¯¯¯¯¯X=

Answer 1

Given that,

mean = = 3.66

standard deviation = = 2.9598

n = 60

= = 3.66

= / n = 2.9598 / 60 = 0.38

Using standard normal table,

P(Z < z) = 90%

= P(Z < z ) = 0.90

= P(Z < 1.282 ) = 0.90  

z = 1.282

Using z-score formula  

= z * +   

= 1.282 * 0.38 + 3.66

= 4.15