Answer :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{160 - 159}{1.68}[/tex]
[tex]Z = 0.6[/tex]
[tex]Z = 0.6[/tex] has a pvalue of 0.7257
X = 150
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{158 - 159}{1.68}[/tex]
[tex]Z = -0.6[/tex]
[tex]Z = -0.6[/tex] 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