2. 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 the mean. b.) What percent of students did better than Jeff? What percent did worse?

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

Step-by-step 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

may i see the picture? it would a lot, xox