Question

Scores of a standardized test are approximately normally distributed with a mean of 85 and a standard deviation of 5.5.
(a) What proportion of the scores is above 90?
(b) What is the 25th percentile of the scores?
(c) If a score is 94, what percentile is it on?

Answer 1

Solution :

Given that,

mean = = 85

standard deviation = =5.5

P(x > 90) = 1 - P(x< 90)

= 1 - P[(x -) / < (90-85) /5.5 ]

= 1 - P(z <0.91 )

Using z table

= 1 -  0.8186

= 0.1814

B

= P(Z < z) = 25%  

= P(Z <-0.67 ) = 0.25

z = -0.67 ( Using standard normal z table,)

Using z-score formula  

x= z * +

x= -0.67*5.5+85

x= 81.3

C

P(x< 94)

= P[(x -) / < (94-85) /5.5 ]

= P(z <1.64 )

=0.9495

=94.95%