Question

For a very large set of data, the measurement mean was found to be 2653 with a standard deviation of 1045. What percentage of the data fall between 1000 and 3000?

Answer 1

For a very large set of data, the measurement mean was found to be 2653 with a standard deviation of 1045. What percentage of the data fall between 1000 and 3000?

Since large set of data , we assume data approximately follows normal distribution.

Z value for 1000, z =(1000-2653)/1045 = -1.58

Z value for 3000, z =(3000-2653)/1045 = 0.33

P( 1000<x<3000) = P( -1.58 <z<0.33)

=P(z < 0.33) – P( z < -1.58)

=0.6293 -0.0571

=0.5722

The required percentage = 57.22%