Question

Some nations require their students to pass an exam before earning their primary school degrees or diplomas. A certain nation gives students an exam whose scores are normally distributed with a mean of 414141 points and a standard deviation of 999 points.
Suppose we select 222 of these testers at random, and define the random variable DDD as the difference between their scores. We can assume that their scores are independent.
Find the probability that their scores are within 101010 points of each other.
You may round your answer to two decimal places.

Answer 1

Using subtraction of normal variables, it is found that there is a 0.5704 = 57.04% probability that their scores are within 10 points of each other.

Normal Probability Distribution

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

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

• It 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, which is the percentile of X.

• When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this problem:

• Two variables from the same population, with mean 41, are subtracted, hence the mean of the distribution of the subtraction is $\mu = 41 - 41 = 0$.

• For each variable, the standard deviation is of 9, hence the standard deviation of the distribution of the subtraction is $\sigma = \sqrt{9^2 + 9^2} = 12.73$

The probability that their scores are within 10 points of each other is the probability that the subtraction is between -10 and 10, that is, the p-value of Z when X = 10 subtracted by the p-value of Z when X = -10.

X = 10:

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

$Z = \frac{10 - 0}{12.73}$

$Z = 0.79$

$Z = 0.79$ has a p-value of 0.7852.

X = -10:

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

$Z = \frac{-10 - 0}{12.73}$

$Z = -0.79$

$Z = -0.79$ has a p-value of 0.2148.

0.7852 - 0.2148 = 0.5704.

0.5704 = 57.04% probability that their scores are within 10 points of each other.

You can learn more about subtraction of normal variables at brainly.com/question/14397951