Act scores for 1,171,460 members of the 2004 high school graduating class who took the test closely followed the normal distribution with mean 20.9 and standard deviation 4.8. Choose two students independently and at random from this group.
A) what is the expected difference in their scores
b) what is the standard deviation in their scores
c) find the probability that the difference in the two students scores is greater than 6.

Thank you so much for your help!!

a) Let x = first score and y = second score

E[x] = 20.9 and E[y] = 20.9

E[x-y] = E[x] – E[y] = 20.9 – 20.9 = 0
b) Standard deviation

= Var[x-y] = Var[x] + Var [y]

= 4.8^2 + 4.8^2 = 46.08

SD[x-y] = sqrt(Var[x-y])

= sqrt(46.08)

= 6.8

c) Z = +/- (mean-x)/SD = +- (0-6)/6.8 = +/- 0.88

From Z table: P(Z<-0.88 or Z>0.88)

= 2*P(Z>0.88)

= 2*0.1894

= 0.3788

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joaobezerra joaobezerra · Answer 2 · 2021-10-15T16:42:55+02:00

Using subtraction of normal variables, it is found that:

a) The expected difference in their scores is 0.

b) The standard deviation in the difference of their scores is 6.7882.

c) 0.3788 = 37.88% probability that the difference in the two students scores is greater than 6.

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

Item a:

Two values from the same distribution, thus the subtraction of the means is 0, which means that the expected difference in their scores is 0.

Item b:

The standard deviation is the square root of the sum of the variances, then:

[tex]\sigma = \sqrt{4.8^2 + 4.8^2} = 6.7882[/tex]

The standard deviation in the difference of their scores is 6.7882.

Item c:

First, we find the z-score when X = 6.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{6 - 0}{6.7882}[/tex]

[tex]Z = 0.88[/tex]

The difference can be of 6 to both sides, thus we want P(|Z| > 0.88), which is 2 multiplied by the p-value of Z = -0.88.
Looking at the z-table, Z = -0.88 has a p-value of 0.1894.
2 x 0.1894 = 0.3788

0.3788 = 37.88% probability that the difference in the two students scores is greater than 6.

A similar problem is given at https://brainly.com/question/24250158