Question

Q scores have a normal distribution with µ = 90 and σ = 10. a) Find the probability for a score over 100. b) Find the score needed for the top 5%

Answer 1

Solution :

Given ,

mean = = 90

standard deviation = = 10

(A)P(x > 100) = 1 - P(x<100 )

= 1 - P[ X - / / (100-90) /10 ]

= 1 - P(z < 1)

Using z table

= 1 - 0.8413

= 0.1587

probability= 0.1587

(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 *10+90

x= 106.4