Question

In a normal distribution of 100, what percent of scores would be expected to find between z=1.30 and z=1.90 ?

Answer 1

`6.8 %`

Explanation:

The Normal Probability Curve has the equation for the area

Area`=int_(z_1)^(z_2)(1/sqrt(2pi))*e^(-.5 z^2)dz`

Area`=int_(1.3)^(1.9)(1/sqrt(2pi))*e^(-.5 z^2)dz`

Area`=0.0680839247`

This area gives the percent. Therefore `6.8 %`

God bless....I hope the explanation is useful.