Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex];
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:
[tex]\mu = 3.66, n = 2.9598, n = 50, s = \frac{2.9598}{\sqrt{50}} = 0.4186[/tex]
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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]0.842 = \frac{X - 3.66}{0.4186}[/tex]
[tex]X - 3.66 = 0.4186*0.842[/tex]
[tex]X = 4.01[/tex]
x¯ = 4.01.
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 [tex]4.012[/tex]
The given parameters are:
[tex]\mu = 3.66[/tex] --- the population mean
[tex]\sigma = 2.9598[/tex] --- the population standard deviation
[tex]n = 50[/tex] ---- the sample size
[tex]P(\bar x \le x) = 80\%[/tex] --- the p value
First, we calculate the sample standard deviation
[tex]\sigma_x = \frac{\sigma}{\sqrt n}[/tex]
This gives:
[tex]\sigma_x = \frac{2.9598}{\sqrt{50}}[/tex]
[tex]\sigma_x = 0.4186[/tex]
The p value is represented as:
[tex]P(\bar x \le x) = P(Z < z)[/tex]
Where:
[tex]z = \frac{x - \mu}{\sigma_x}[/tex]
This gives
[tex]z = \frac{\bar x - 3.66}{0.4186}[/tex]
Substitute [tex]P(\bar x \le x) = 80\%[/tex] in [tex]P(\bar x \le x) = P(Z < z)[/tex]
[tex]80\% = P(Z < z)[/tex]
Substitute [tex]z = \frac{\bar x - 3.66}{0.4186}[/tex] in [tex]80\% = P(Z < z)[/tex]
[tex]80\% = P(Z < \frac{\bar x - 3.66}{0.4186})[/tex]
The z value, when [tex]p = 80\%[/tex] is:
[tex]z = 0.84162[/tex]
So, we have:
[tex]0.84162 = \frac{\bar x - 3.66}{0.4186}[/tex]
Multiply both sides by 0.4186
[tex]0.4186 \times 0.84162 = \bar x - 3.66[/tex]
[tex]0.352302132 =\bar x - 3.66[/tex]
Solve for x
[tex]\bar x = 0.352302132 +3.66[/tex]
[tex]\bar x = 4.012302132[/tex]
Approximate
[tex]\bar x = 4.012[/tex]
Hence, the mean of pups born is [tex]4.012[/tex]
Read more about probability from z score at:
https://brainly.com/question/3827012
