Question

For of a bell-shaped data with mean of 115 and standard deviation of 22, approximately

a) 0.15% of the values lies below:

b) 68% of the middle values lies between:  and

c) 2.5% of the values lies above:

d) 0.15% of the values lies above:

Answer 1

Solution :

Given that,

mean = = 115

standard deviation = = 22

Using standard normal table,

a ) P( Z < z) =0.15%
P(Z < z) = 0.0015

z =-2.97

Using z-score formula,

x = z * +

= - 2.97 *22 + 115

= 49.66

x = 49.66

P(-z < Z < z) = 68%
P(Z < z) - P(Z < z) = 0.68
2P(Z < z) - 1 = 0.68
2P(Z < z ) = 1 + 0.68
2P(Z < z) = 1.68
P(Z < z) = 1.68 / 2
P(Z < z) = 0.84
z = 0.99and z = - 0.99

Using z-score formula,

x = z * +

= 0.99*22+115

= 136.78

x = 136.78

x = z * +

=-0.99*22+115

= 93.22

x = 93.22

c ) P( Z > z) =2.5%

P(Z > z) = 0.025

1 - P( Z < z) = 0.025

P(Z < z) = 1 - 0.025

P(Z < z) = 0.975

z = 1.96

Using z-score formula,

x = z * +

= 1.96*22+115

= 158.12

x = 158.12

d ) P( Z > z) =0.15%

P(Z > z) = 0.0015

1 - P( Z < z) = 0.0015

P(Z < z) = 1 - 0.0015

P(Z < z) = 0.9985

z = 2.97

Using z-score formula,

x = z * +

= 2.97*22+115

= 180.34

x = 180.34