Question

Test scores from a college math course follow a normal distribution with mean = 72 and standard deviation = 8

Let x be the test score. Find the probability for a) P(x < 66)
b) P(68<x<78)
c) P(x>84)

d) If 600 students took this test, how many students scored between 62 and 72?

Answer 1

Solution :

Given that,

mean = = 72

standard deviation = = 8

a ) P( x < 66 )

P ( x - / ) < ( 66 - 72 /8)

P ( z < -6 / 8 )

P ( z < -0.75)

= 0.2266

Probability =0.2266

b ) P (68< x < 78 )

P ( 68 - 72 / 8) < ( x -  / ) < ( 78 - 72 / 8)

P ( - 4 / 8 < z < 6 / 8 )

P (-0.5 < z < 0.75)

P ( z < 0.75 ) - P ( z < -0.5)

Using z table

=0.7734 - 0.3085

= 0.4649

Probability = 0.4649

c ) P (x > 84 )

= 1 - P (x < 84 )

= 1 - P ( x -  / ) < ( 84 - 72 / 8)

= 1 - P ( z < 12 / 8 )

= 1 - P ( z < 1.5 )

Using z table

= 1 - 0.9332

= 0.0668

Probability = 0.0668

d ) N =600

P (62< x < 72 )

P ( 62 - 72 / 8) < ( x -  / ) < (72 - 72 / 8)

P ( - 10 / 8 < z < 0 / 8 )

P (-1.25 < z < 0)

P ( z < 0 ) - P ( z < -1.25)

Using z table

= 0.5000 - 0.1056

= 0.3944

Probability = 0.3944

600 * 0.3944 =237