Question

A set of final examination grades in an introductory statistics course is normally distributed, with a mean of 73 and a standard deviation of 8.2.
(i) What is the probability that a student scored below 91 on this exam?
(ii) What is the probability that a student scored between 65 and 89?
(iii) Given that 15% of the students scored grade B. What is the cut-off mark for grade B?

Answer 1

Solution :

Given that ,

mean = = 73

standard deviation = = 8.2

i) P(x < 91 ) = P[(x - ) / < (91 -73) /8.2 ]

= P(z < 2.20 )

Using standard noramal table

= 0.9861

probability = 0.9861

ii)

P( 65 < x < 89 ) = P[(65 -73)/8.2 ) < (x - ) /  < (89 -73) / 8.2) ]

= P( -0.98< z < 1.95)

= P(z < 1.95 ) - P(z < -0.98)

= 0.9744 - 0.1635 = 0.8109

Probability = 0.8109

iii) 15% is

P(Z < z ) = 0.15

z = -1.036

Using z-score formula,

x = z * +

x = -1.036 * 8.2 + 73

x = 64.50

Answer = 64.50 B grade marks.