Question

The grades of an examination whose mean is 72 and whose standard deviation is 12 are normally distributed.

A.) Anyone who scores below 55 will be retested. What percentage does this represent?

B.) Those who score in the top 5% are to receive a special commendation. What score must be surpassed to receive this special commendation?

Answer 1

Solution :

Given that ,

mean =   = 72

standard deviation = σ   = 12

P(X< 55) = P[(X- ) / σ   < (55-72) /12 ]

= P(z <-1.42 )

Using z table

= 0.0778

probability =7.78%

(B)

Using standard normal table,

P(Z > z) = 5%

= 1 - P(Z < z) = 0.05

= P(Z < z ) = 1 - 0.05

= P(Z < z ) = 0.95

= P(Z < 1.64 ) = 0.95

z = 1.64 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x= 1.64 *12+72

x= 91.68

x=92