Answer:
a.) The z-score of Jeff's SAT score is z=-1.75.
This means Jeff's score is 1.75 standard deviations below or away from the mean
b)
What percent of students did better than Jeff?
= 96% of the students did worse
What percent did worse?
= 4 % of the students did worse
Explanation:
Consider SAT scores for 2009 high school graduates which are approximately normal with a mean of 1050 and a standard deviation of 150. a.) The z-score of Jeff's SAT score is z=-1.75.
This means Jeff's score is 1.75 standard deviations away from the mean.
b.)
The z-score of Jeff's SAT score is z=-1.75.
We have to find the percentile that Jeff is in by finding the probability of his z score using the z table
P(x<Z) = P(z = -1.75)
= 0.040059
Converting to percentage
= 0.040059 × 100
= 4.0059%
Approximately = 4%
Hence, Jeff is in the 4th percentile
This means Jeff scored higher than 4% of his colleagues
What percent of students did better than Jeff?
= 100 - 4%
= 96% of the students did better than Jeff.
What percent did worse?
4% of the students did worse