Question

1)In a distribution of IQ scores, where the mean is 100 and the standard deviation is 15........

-Compute the z score for an IQ of 100

-Compute the z score for an IQ of 107

2)For all US women, assuming a normal distribution - Mean height is 64 inches ; Standard deviation is 2.4 inches

-What percentage of US women are 60 inches or shorter?

-What percentage of US women have a height between 64 and 67 inches?

Answer 1

Solution :

Given that ,

1mean = = 100

standard deviation =  = 15

a) x = 100

Using z-score formula,

z = x - /   

z = 100 - 100 / 15

z = 0

b) x = 107

Using z-score formula,

z = x - /   

z = 107 - 100 / 15

z = 0.47

2) Given that ,

mean = = 64 in.

standard deviation = = 2.4 in.

a) P(x 60)

= P[(x - ) / (60 - 64) / 2.4]

= P(z -1.60)

Using z table,

= 0.0548

The percentage is = 5.48%

b) P(64 < x < 67) = P[(64 - 64)/2.4 ) < (x - ) /  < (67 - 64) / 2.5) ]

= P(0 < z < 1.25)

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

Using z table,

= 0.8944 - 0.5

=0.3944

The percentage is = 39.44%