The mean points obtained in an aptitude examination is 103 points with a variance of 169. What is the probability that the mean of the sample would differ from the population mean by less than 2.8 points if 63 exams are sampled? Round your answer to four decimal places.

0.9128 = 91.28% probability that the mean of the sample would differ from the population mean by less than 2.8 points if 63 exams are sampled

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 and standard deviation(which is the square root of the variance) , the zscore 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 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 random variable X, with mean and standard deviation , the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

What is the probability that the mean of the sample would differ from the population mean by less than 2.8 points if 63 exams are sampled?

This is the pvalue of Z when X = 103+2.8 = 105.8 subtracted by the pvalue of Z when X = 103 - 2.8 = 100.2. So

X = 105.8

By the Central Limit Theorem

has a pvalue of 0.9564

X = 100.2

has a pvalue of 0.0436

0.9564 - 0.0436 = 0.9128

0.9128 = 91.28% probability that the mean of the sample would differ from the population mean by less than 2.8 points if 63 exams are sampled