(1 point)
The number of eggs that a female house fly lays during her lifetime is normally distributed with mean 790 and standard deviation 92.
Random samples of size 98 are drawn from this population, and the mean of each sample is determined. What is the probability that
the mean number of eggs laid would differ from 790 by less than 30? Round your answer to four decimal places.

Answer:

0.9988 = 99.88% probability that the mean number of eggs laid would differ from 790 by less than 30.

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 normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, 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 variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean 790 and standard deviation 92.

This means that [tex]\mu = 790, \sigma = 92[/tex]

Samples of 98

This means that [tex]n = 98, s = \frac{92}{\sqrt{98}}[/tex]

What is the probability that the mean number of eggs laid would differ from 790 by less than 30?

This is the pvalue of Z when X = 790 + 30 = 820 subtracted by the pvalue of Z when X = 790 - 30 = 760. So

X = 820

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{820 - 790}{\frac{92}{\sqrt{98}}}[/tex]

[tex]Z = 3.23[/tex]

[tex]Z = 3.23[/tex] has a pvalue of 0.9994

X = 760

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{760 - 790}{\frac{92}{\sqrt{98}}}[/tex]

[tex]Z = -3.23[/tex]

[tex]Z = -3.23[/tex] has a pvalue of 0.0006

0.9994 - 0.0006 = 0.9988

0.9988 = 99.88% probability that the mean number of eggs laid would differ from 790 by less than 30.

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