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 
and standard deviation 
, the z-score 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 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 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 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 
Samples of 98
This means that 
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
By the Central Limit Theorem
has a pvalue of 0.9994
X = 760
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.
