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 = 3.66 μ=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 80 % 80% of all random samples of such squirrels of size n = 50 , n=50, the mean number of pups born to females in the sample is less than ¯¯¯ x .

The number of squirrel pups follows a non-normal distribution.

The mean number of squirrel pups born out of a sample size of 50 is

The given parameters are:

--- the population mean

--- the population standard deviation

---- the sample size

--- the p value

First, we calculate the sample standard deviation

This gives:

The p value is represented as:

Where:

This gives

Substitute in

Substitute in

The z value, when is:

So, we have:

Multiply both sides by 0.4186

Solve for x

Approximate

Hence, the mean of pups born is

Read more about probability from z score at:

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x¯ = 4.01.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation ;

For a skewed(non-normal) variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

Determine the mean number of pups, ¯¯¯ x , x¯, such that in 80 % 80% of all random samples of such squirrels of size n = 50 , n=50, the mean number of pups born to females in the sample is less than ¯¯¯ x .

This is x when Z has a pvalue of 0.8. So X when Z = 0.842.

By the Central Limit Theorem

x¯ = 4.01.