Question

The area between:

Z=-0.64 and Z=0.64

Z= -0.66 and Z=0

Z= -2.14 and Z= -0.74

Find the Z-scores that separate the middle 66% of the distribution from the area in the tails of the standard normal distribution.

Answer 1

1) P(-0.64<Z<0.64)=P(Z>-0.64)-P(Z>0.64)

=1-P(Z<-0.64)-P(Z>0.64)

=1-P(Z>0.64)-P(Z>0.64)

=1-2*P(Z>0.64)

=1-2*0.26109

=0.4778

2) P(-0.66<Z<0)=P(Z>-0.66)-P(Z>0)

=1-P(Z<-0.66)-P(Z>0)

=1-P(Z>0.66)-P(Z>0)

=1-0.25463-0.5

=0.2454

3) P(-2.14<Z<-0.74)=P(Z>-2.14)-P(Z>-0.74)

=[1-P(Z<-2.14)]-[1-P(Z<-0.74)]

=1-P(Z>2.14)-1+P(Z>0.74)

=P(Z>0.74)-P(Z>2.14)

=0.22965-0.016177

=0.2135

4) Let , P(-a<Z<a)=0.66

From normal probability table ,

a=0.95

Therefore , the Z-scores -0.95 and 0.95 that separate the middle 66% of the distribution from the area in the tails of the standard normal distribution.