Question

g The average midterm score of students in a certain course is 70 points. From the past experience it is known that the midterm scores in this course are Normally distributed. If 29 students are randomly selected and the standard deviation of their scores is found to be 13.15 points, find the probability that the average midterm score of these students is at most 75 points. (Round your final answer to 3 places after the decimal point).

Answer 1

Answer:

0.98 = 98% probability that the average midterm score of these students is at most 75 points.

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The average midterm score of students in a certain course is 70 points.

This means that $\mu = 70$

29 students are randomly selected and the standard deviation of their scores is found to be 13.15 points.

This means that $\sigma = 13.15, n = 29, s = \frac{13.15}{\sqrt{29}} = 2.44$

Find the probability that the average midterm score of these students is at most 75 points.

This is the pvalue of Z when X = 75. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{75 - 70}{2.44}$

$Z = 2.05$

$Z = 2.05$ has a pvalue of 0.98.

0.98 = 98% probability that the average midterm score of these students is at most 75 points.