Question

A grading scale is set up for 1000 students’ test scores. It is assumed that the scores are normally distributed with a mean score of 80 and a standard deviation of 10:

a)    What proportion of students will have scores between 40 and 85?

b)    If 60 is the lowest passing score, what proportion of students pass the test?

c)    What score would a student have to score to be in the 68^th percentile?

d)    What score would a student have to make to be in the top 20% of the class?

e) If 60 is the lowest passing score, estimate how many students pass the test?

Answer 1

Solution :

Given that ,

mean = = 80

standard deviation = = 10

a) P( 40< x < 85 ) = P[(40 -80)/10 ) < (x - ) /  < (85 -80) /10 ) ]

= P( -4< z <0.5 )

= P(z < 0.5 ) - P(z < -4 )

Using standard normal table

= 0.6915 - 0 = 0.6915

Proporation = 0.6915

2)

P(x < 60 ) = P[(x - ) / < (60 -80) /10 ]

= P(z < -2 )

= 0.0228

proporation =0.0288

3) 68 th percentile

P(Z < z) = 0.68

z = 0.468

Using z-score formula,

x = z * +

x = 0.468 * 10 +80

x = 84.68

d)Top 20% = 0.20

P(Z > z ) = 0.20

1- P(z < z) =0.20

P(z < z) = 1-0.20 = 0.80

z = 0.84

Using z-score formula,

x = z * +

x = -0.84*10+80

x = 88.4

e)

P(Z < z ) = 0.60

z = 0.25

Using z-score formula,

x = z * +

x =0.25 *10 +80

x = 82.5 = 83

Answer = 83 student