Question

At the prestigious university, they found their mean of the 30 accepted students was in the 98^thpercentile of all sample means. Individual scores, X, are normally distributed with a mean of 550 and a standard deviation of 100.

a.) What was their mean for the 30 observations?  

b.) What is the probability of getting a mean of 30 students that scored higher than 685?

Answer 1

Solution :

mean = = 550

standard deviation = =100

n = 30

= 550

₌ / n = 100 30 = 37.94733

Using standard normal table,

a ) P(Z < z) = 98%

P(Z < z) = 0.98

P(Z < 2.054 ) = 0.98

z = 2.05

Using z-score formula,

= z * +

= 2.05 * 18.2574 + 550

= 587.42

= 587.42

b ) P ( X >  685 )

= 1 - P ( X < 685 )

= 1 - P ( X - /) < (685 - 550 / 100 )

= 1 - P( z < 135 / 100 )

= 1 - P ( z < 1.35 )   

Using z table

= 1 - 0.9115

= 0.0885

Probability = 0.0885