The college Physical Education Department offered an Advanced First Aid course last summer. The scores on the comprehensive final exam were normally distributed, and the z scores for some of the students are shown below.
Robert, 0.96
Juan, 1.78
Susan, –2.2
Joel, 0.00
Jan, –0.89
Linda, 1.45
(a) Which of these students scored above the mean? (Select all that apply.)
Jan
Joel
Juan
Linda
Robert
Susan
(b) Which of these students scored on the mean? (Select all that apply.)
Jan
Joel
Juan
Linda
Robert
Susan
(c) Which of these students scored below the mean? (Select all that apply.)
Jan
Joel
Juan
Linda
Robert
Susan
(d) If the mean score was ? = 156 with standard deviation ? = 24, what was the final exam score for each student? (Round your answers to the nearest whole number.)
Robert-
Joel-
Jan-
Juan-
Susan-
Linda-

The standardized normal distribution N (0,1) that expresion stands for normal distribution with mean 0 and standard deviation 1, therefore all positive values are above 0 and all negativa values are below 0

The sample n is 6 then we should use t-student distribution, we calculate t statistic as follow

t = (x-μ)/εε where standard error is εε = σ/√6 εε = 9.795 since σ =24

Robert

0.96 * 9.795 = x - 156 ⇒ x = 165.4

1.78 * 9.795 = x - 156 x = 173.4

-2.2 * 9.795 = x - 156 x = 134.5

0 * 9.795 = x- 156 x= 156

-0.89* 9.795 = x -156 x = 147.3

1.45 * 0.9795 = x -156 x = 170.2