Question

Assume a normal distribution with a mean of 90 and a standard deviation of 7. what limits would include the middle 75% of the cases.

Answer 1

Solution:

Given:

A random variable X follows Normal l distribution with a mean of 90 and a standard deviation of 7.

That is: X ~ Normal

We have to find limits that would include the middle 75% of the cases.

That is find x1 and x2 such that:

P(x1 < X < x2 ) =0.75

Thus find z values such that:

P( -z < Z < z ) = 0.75

Area between -z and +z is 0.75

then area in left tail would be: (1-0.75) / 2 = 0.25 / 2 = 0.1250

That is: area below -z is 0.1250

Thus

Total area below +z would be = ( Area below -z )   + ( Area between -z and z )

Total area below +z would be = 0.1250 + 0.7500

Total area below +z would be = 0.8750

Thus look in z table for Area = 0.8750 or its closest area and find z value.

Area 0.8749 is closest to 0.8750 and it corresponds to 1.1 and 0.05

Thus z = 1.15

Since Normal distribution is symmetric, z value for left tail is same but sign is negative.

That is z value for left tail = -1.15

Now we use following formula to find x values:

x1 is lower limit , hence we use z = -1.15

and

x2 is upper limit , hence we use z = 1.15

Thus the middle 75% of the cases are between 81.95 and 98.05 .