Find proportion given mean and standard diviation

Question

If a student A scored 34 on a test of reading and the test had a mean of 51 and a standard deviation of 9, what proportion of test respondents scored below student A?

Answer 1

Evaluate (34 − 51) Divided By 9.

Obtain -17/9 or -1.888888888... which is the z-value

or argument used to find the proportion of area under the

Standard Normal Curve involved in this problem.

This proportion of area (also known as the normal probability)

is given to high accuracy by a calculator program

as -0.4705466407.

The proportion of tested students with scores lower than

the score for Student A is given by 1 − [|-0.4705466407| + 0.5]

which reduces to 0.0294533593.

Using a printed table, one would locate the argument 1.89 and

trace to the proportion of 0.4706; he would then calculate

1 − |-0.4706| − 0.5 to obtain 0.0294.