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.
