Area under z table

Suppose we know that the height of students of a particular age group are normally distributed with mean 60 inches and standard deviation 6 inches, how do we find the probability that a student randomly selected has height less than 50 inches?

  • The first step is to convert the given normal distribution to standard normal distribution (Standard normal distribution has mean 0 and std. devn of 1)
  • Conversion:  Zcorresponding to 50 inches = (50 – mean)/std. devn.
  • Z corresponding to 50 inches = (50 – 60) / 6 = -1.67
  • Now the problem statement can be rewritten as p (Z<= – 1.67)
  • The above probability can be found by looking up the area under the standard normal curve that is to the left of – 1.67
  • How to lookup areas under the curve from the table?

Objective:

Finding the area to the left of -1.67 in standard normal curve.

Suppose, we are given the negative z table with area to the left

Normalcurve3.jpg

normaltable3.jpg

As we require area to the left of -1.67, we can directly use this table and lookup for -1.67 as shown below.

Normallookup3.jpg

Hence the required area is 0.0475 which is p (Z<= – 1.67). Hence, the chance of finding a student with height less than 50 inches is 4.75%..

Suppose, we are given the negative z table with area to the right

Normalcurve4.jpg

Normaltable4.jpg

As we require area to the left of -1.67, we can use this table and lookup for -1.67 as shown below to get the area to the right

Normallookup4.jpg

p (Z>= – 1.67), area to the right of -1.67 is 0.9525

As we require area to the left, we simply subtract the above area from 1

p (Z<= – 1.67) = 1-0.9525 = 0.0475 (4.75% chance)

Suppose, we are given the positive z table with area to the right

Normaltable2.jpg

As we require area to the left of -1.67, we can use this table and lookup for + 1.67 as shown below to get the area to the right

Area to the right of +1.67 is same as the area to the left of -1.67 (As normal curve is symmetrical)

Hence, the required area from above table is 0.0475

Suppose, we are given the positive z table with area to the left

Normaltable1.jpg

Lookup for the area corresponding to +1.67 in the above table and subtract from 1 to get the area to the right of + 1.67.

This area is same as the area to the left of -1.67.

Hence, the required area from above table is 0.0475.

We can infer for any area corresponding to a z value if any one of the above tables is provided.