Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100. Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5. Who performed better on the standardized exams and why? Reagan scored higher than Jessie. Reagan's score was 1140, which is greater than Jessie's score of 30. Reagan scored higher than Jessie. Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean, but closer to the mean than Jessie's standardized score of 2.6 standard deviations above the mean. Reagan scored higher than Jessie. Reagan's score was 140 points above the mean of 1000, and Jessie's was 13 points above the mean of 17. Jessie scored higher than Reagan. Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean. Jessie scored higher than Reagan. Jessie is only 6 points from the top score of 36 on the ACT, and Reagan is 460 points from the top score of 1600 on the SAT.

We are given that Reagan scored 1140 on the SAT. The distribution of SAT scores in a reference population is normally distributed with mean 1000 and standard deviation 100.

Jessie scored 30 on the ACT. The distribution of ACT scores in a reference population is normally distributed with mean 17 and standard deviation 5.

For finding who performed better on the standardized exams, we have to calculate the z-scores for both people.

1. Finding z-score for Reagan;

Let X = distribution of SAT scores

SO, X ~ Normal([tex]\mu=1000, \sigma^{2}=100^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 1000

[tex]\sigma[/tex] = standard deviation = 100

Now, Reagan scored 1140 on the SAT, that is;

z-score = [tex]\frac{1140-1000}{100}[/tex] = 1.4

2. Finding z-score for Jessie;

Let X = distribution of ACT scores

SO, X ~ Normal([tex]\mu=17, \sigma^{2}=5^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean = 17

[tex]\sigma[/tex] = standard deviation = 5

Now, Jessie scored 30 on the ACT, that is;

z-score = [tex]\frac{30-17}{5}[/tex] = 2.6

This means that Jessie scored higher than Reagan because Jessie's standardized score was 2.6, which is 2.6 standard deviations above the mean and Reagan's standardized score was 1.4, which is 1.4 standard deviations above the mean.