Problem 57E

Coffee Contents During a quality assurance check, the actual coffe contents (in ounces) of six jars of instant coffee were recorded as 6.03, 5.59, 6.40, 6.00, 5.99, and 6.02.

(a) Find the mean and the median of the coffee content.

(b) The third value was incorrectly measured and is actually 6.04. Find the mean and the median of the coffee content again.

(c) Which measure of central tendency, the mean or the median, was affected more by the data entry error?

Solution

Step 1 of 3

Given that the coffee content of 6 jars were recorded as follows

6.03, 5.59, 6.40, 6.00, 5.99, 6.02

a) We have to find the mean and median of the coffee content

Means is the average of all the values

Mean=

=

=36.03/6

=6.005

Median is the middle value of the data set

To find median first arrange the values in the ascending order

5.59, 5.99, 6.00, 6.02, 6.03, 6.40

Median =

=(6+1)/2 th value

=(3.5)th value

=(3rd value+4th value)/2

=(6.00+6.02)/2

=6.01

Hence Mean=6.005

Median=6.01