During a quality assurance check the actual coffee content i

During a quality assurance check, the actual coffee content (in ounces) of six jars (a) The mean coffee content is ounces. of instant coffee was recorded as 6.03, 5.61, 6.50, 6.00, 5.99, and 6.02. (Round to three decimal places as needed) (a) Find the mean and the median of the coffee content. The median coffee content is ounces (Round to three decimal places as needed.) (b) The third value was incorrectly measured and is actually 6.05. Find the mean and median of the coffee content again. (b) After correcting the data error, the mean coffee content is ounces. (c) Which measure of central tendency, the mean or the median, was affected (Round to three decimal places as needed.) more by the data entry error? After correcting the data error, the median coffee content is ounces (Round to three decimal places as needed.) (c) Which measure of central tendency, the mean or the median, was affect more by the data entry error? Median Mean

Solution

6.03 , 5.61 , 6.50 , 6.00 , 5.99 , 6.02

Mean = (6.03 + 5.61 + 6.50 + 6.00 + 5.99 + 6.02) / 6

Mean = 6.025 --> FIRST ANSWER of part a

Median : So, first lets arrange in increasing order....
5.61 , 5.99 , 6.00 , 6.02 , 6.03 , 6.50

Here we have a total of 6 values

So, median = (3rd + 4th) / 2

median = (6.00 + 6.02) / 2

median = 6.01 ----> SECOND ANSWER of part a

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Mean = (6.03 + 5.61 + 6.05 + 6.00 + 5.99 + 6.02) / 6

Mean = 5.95 ---> FIRST ANSWER for part b

Median : Increasing order :

5.61 , 5.99 , 6.00 , 6.02 , 6.03 , 6.05

Median = (3rd + 4th) / 2 = (6.00 + 6.02) / 2 ---> 6.01 ---> SECOND ANSWER for part b

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Which was affected :

Mean ----> ANSWER

Reason : The mean in part a was 6.025 but in part b was 5.95 whereas, the median was unchanged in both parts