Question

Suppose the grade on a Math test is normally distributed with mean 78 and standard deviation 10. (a) Compute the z-scores (5 points) (a-1) If Bob got 70 on the test, what is his z-score? (a-2) If Jane got 90 on the test, what is her z-score? (b) Compute the actual grades (5 points) (b-1) Suppose David achieved a grade 1.8 standard deviation above the mean (? = 1.8), what was his actual grade? (b-2) Suppose Lily achieved a grade 0.5 standard deviation below the mean (? = −0.5), what was her actual grade? (c) Rob achieved a grade that exceeded 95% of all grades. Find Rob’s actual grade. (6 points) (d) Suppose 32% of students did better than Mei. Find Mei’s actual grade. (6 points)

Answer 1

(a)

(a -1)

Z = (X- )/

= (70 - 78)/10 = - 0.80

So,

Answer is:

- 0.80

(a -2)

Z = (X- )/

= (90 - 78)/10 = 1.20

So,

Answer is:

1.20

(b)

(b -1)

X = 78 + (1.8 X 10) = 78 + 18 = 96

So,

Answer is:

96

(b -2)

X = 78 + (- 0.5 X 10) = 78 - 5 = 73

So,

Answer is:

73

(c)

grade that exceeded 95% of all grades corresponds to area = 0.95 - 0.50 = 0.45 from mid value to Z on RHS.

Table of Area Under Standard Normal Curve gives Z = 1.645

So,

Z = 1.645 = (X - 78)/10

So,

X = 78 + (1.645 X 10) = 94.45

So

Answer is:

94.45

(d)

32% better is equivalent to area = 0.50- 0.32 = 0.18.

Table gives Z = 0.47

So,

X = 78 + (0.47 X 10) = 82.7

So

Answer is:

82.7