Question

Suppose 2500 students take a college exam.The scores have a normal distribution with mean

μ = 52 points and standard deviation σ = 11 points.Estimate how many students scored between
30 and 74 points. (Put in the number only)

For the same distribution of problem 1

a)Estimate what percent of students scored 63 points or more.

b)For a score of 77, what is the z-value (standardized value)?

(Enter the answer as a,b with no % symbol. Round the percent to the nearest one and the z-value to the nearest tenth for example: 26,2.1)

Answer 1

1))

Given :- = 52 , = 11 )
We convet this to Standard Normal as
P(X < x) = P( Z < ( X - ) / )
P ( 30 < X < 74 ) = P ( Z < ( 74 - 52 ) / 11 ) - P ( Z < ( 30 - 52 ) / 11 )
= P ( Z < 2) - P ( Z < -2 )
= 0.9772 - 0.0228
= 0.9545

Of the 2500 students we expect, 2500 * 0.9545 = 2386

a)

X ~ N ( µ = 52 , σ = 11 )
We covert this to standard normal as
P ( X < x) = P ( (Z < X - µ ) / σ )
P ( X > 63 ) = P(Z > (63 - 52 ) / 11 )
= P ( Z > 1 )
= 1 - P ( Z < 1 )
= 1 - 0.8413
= 0.1587

= 15.87%

b)

z-score = ( x - ) /

= (77 - 52) / 11

= 2.27