Question

For the next four questions, assume the following information about a population. The variance (?2) = 9 and the mean (?) = 30. Based on calculation of the z scores and use of the Standard Normal Curve:

What proportion of the scores can be expected to be above 33?

What proportion of the scores can be expected to be between 33 and 36?

What proportion of the scores can be expected to be between 26 and 35?

What are the .95 probable limits of the above set of scores?

Answer 1

Here we are given the distribution as:

a) The proportion is computed here as:

P( X > 33)

Converting this to a standard normal variable, we get here:

Getting it from the standard normal tables, we get:

Therefore 0.1587 is the required probability here.

b) The proportion here is computed as:

P( 33 < X < 36 )

Converting it to a standard normal variable, we get:

P( 1 < Z < 2 )

= P(Z < 2) - P(Z < 1)

Converting to a standard normal tables, we get:

= 0.9772 - 0.8413

= 0.1359

Therefore 0.1359 is the required proportion here.

c) The proportion here is computed as:

P( 26 < X < 35 )

Converting it to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.8610 is the required probability here.

d) From standard normal tables, we have:

P( -1.96 < Z < 1.96 ) = 0.95

Therefore the lower and upper limits here are computed as:
= Mean - 1.96*Std Dev
= 30 - 1.96*3
= 24.12

Upper limit = 30 + 5.88 = 35.88

Therefore limits here are ( 24.12, 35.88 )