Mathematics, 24.03.2020 00:32 jessv01

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 .

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Answer from: chartrow7

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 4.012

The given parameters are:

$\mu = 3.66$ --- the population mean

$\sigma = 2.9598$ --- the population standard deviation

n = 50 ---- the sample size

$P(\bar x \le x) = 80\%$ --- the p value

First, we calculate the sample standard deviation

$\sigma_x = \frac{\sigma}{\sqrt n}$

This gives:

$\sigma_x = \frac{2.9598}{\sqrt{50}}$

$\sigma_x = 0.4186$

The p value is represented as:

$P(\bar x \le x) = P(Z < z)$

Where:

$z = \frac{x - \mu}{\sigma_x}$

This gives

$z = \frac{\bar x - 3.66}{0.4186}$

Substitute $P(\bar x \le x) = 80\%$ in $P(\bar x \le x) = P(Z < z)$

$80\% = P(Z < z)$

Substitute $z = \frac{\bar x - 3.66}{0.4186}$ in $80\% = P(Z < z)$

$80\% = P(Z < \frac{\bar x - 3.66}{0.4186})$

The z value, when $p = 80\%$ is:

z = 0.84162

So, we have:

$0.84162 = \frac{\bar x - 3.66}{0.4186}$

Multiply both sides by 0.4186

$0.4186 \times 0.84162 = \bar x - 3.66$

$0.352302132 =\bar x - 3.66$

Solve for x

$\bar x = 0.352302132 +3.66$

$\bar x = 4.012302132$

Approximate

$\bar x = 4.012$

Hence, the mean of pups born is 4.012

Read more about probability from z score at:

link

Answer from: quanwalker651370

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ ;

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:

$\mu = 3.66, n = 2.9598, n = 50, s = \frac{2.9598}{\sqrt{50}} = 0.4186$

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.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$