Question

With a height of 61 inches, Alexander was the shortest president of a particular club in the past century. The club presidents of the past century have a mean height of 62.1 inches and a standard deviation of 2.8 in.

a. What is a positive difference between ALexanders height and the mean?

b. How many standard deviations is that [the difference found in part (a)]?

c. Convert Alexanders height to a z scores

d. if we consider "usual" heights to be those that convert to z scores between -2 and 2, is alexanders height usual or unusual?

a. The positive difference between alexanders height and the mean is _ in.

Answer 1

Solution:

We are given

Alexander’s height = 61 inches

Mean = 62.1 inches

Standard deviation = SD = 2.8 inches

a. What is a positive difference between ALexanders height and the mean?

Required difference = |Alexander’s height – mean| = |61 – 62.1| = | -1.1| = 1.1

Required difference = 1.1 inch

b. How many standard deviations is that [the difference found in part (a)]?

We have difference = 1.1, SD = 2.8

Required number of standard deviations = 1.1/2.8 = 0.392857

Answer: 0.392857

c. Convert Alexanders height to a z scores

Z = (X – mean) / SD

Z = (61 – 62.1)/2.8 = -1.1/2.8 = -0.39286

Z = -0.39286

d. if we consider "usual" heights to be those that convert to z scores between -2 and 2, is alexanders height usual or unusual?

Alexander’s height is usual because Z = -0.39286 is lies between -2 and 2.