Answer:
0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled
Step-by-step explanation:
To solve this question, we have 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
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
, a large sample size can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68](https://tex.z-dn.net/?f=%5Cmu%20%3D%20159%2C%20%5Csigma%20%3D%2013%2C%20n%20%3D%2060%2C%20s%20%3D%20%5Cfrac%7B13%7D%7B%5Csqrt%7B60%7D%7D%20%3D%201.68)
What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?
This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So
X = 160
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{160 - 159}{1.68}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B160%20-%20159%7D%7B1.68%7D)
![Z = 0.6](https://tex.z-dn.net/?f=Z%20%3D%200.6)
has a pvalue of 0.7257
X = 150
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{158 - 159}{1.68}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B158%20-%20159%7D%7B1.68%7D)
![Z = -0.6](https://tex.z-dn.net/?f=Z%20%3D%20-0.6)
has a pvalue of 0.2743
0.7257 - 0.2743 = 0.4514
0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled