Question

The mean points obtained in an aptitude examination is 167 points with a standard deviation of 20 points. What is the probability that the mean of the sample would differ from the population mean by less than 3.8 points if 76 exams are sampled

Answer 1

Answer:

0.9029 = 90.29% probability that the mean of the sample would differ from the population mean by less than 3.8 points if 76 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 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 establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

The mean points obtained in an aptitude examination is 167 points with a standard deviation of 20 points

This means that $\mu = 167, \sigma = 20$

Sample of 76:

This means that $n = 76, s = \frac{20}{\sqrt{76}}$

What is the probability that the mean of the sample would differ from the population mean by less than 3.8 points?

P-value of Z when X = 167 + 3.8 = 170.8 subtracted by the p-value of Z when X = 167 - 3.8 = 163.2. So

X = 170.8

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

By the Central Limit Theorem

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

$Z = \frac{170.8 - 167}{\frac{20}{\sqrt{76}}}$

$Z = 1.66$

$Z = 1.66$ has a p-value of 0.9515

X = 163.2

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

$Z = \frac{163.2 - 167}{\frac{20}{\sqrt{76}}}$

$Z = -1.66$

$Z = -1.66$ has a p-value of 0.0485

0.9514 - 0.0485 = 0.9029

0.9029 = 90.29% probability that the mean of the sample would differ from the population mean by less than 3.8 points if 76 exams are sampled