During a quality assurance check, the actual contents (in grams) of six containers of protein powder were recorded as 1530, 1532, 1495, 1508, 1528, and 1511. (a) Find the mean and the median of the contents. (b) The third value was incorrectly measured and is actually 1515. Find the mean and the median of the contents again. (c) Which measure of central tendency, the mean or the median, was affected more by the data entry error?

Step-by-step explanation:

Given the values of the actual content of protein powder recorded as shown:

1530, 1532, 1495, 1508, 1528, and 1511

a) We are to find the mean and median of the contents.

Mean is the average sum of the numbers. It is expressed mathematically as xbar = ΣXi/N

Xi are individual values

N is the total number of values present.

N = 6

xbar = (1,530+1,532+1,495 +1,508+1,528+1,511)/6

xbar = 9104/6

xbar = 1517.33

Median of the data is the value at the middle after rearrangement. On rearranging from lowest to highest:

1,495, 1508, 1511, 1528, 1530, 1532

The two values at the centre are 1511 and 1528.

Median = 1511+1528/2

Median = 1519.5

b) If the third value was incorrectly measured and is actually 1515, then our new data will become.

1530, 1532, 1515, 1508, 1528, and 1511

xbar = (1,530+1,532+1,515 +1,508+1,528+1,511)/6

xbar = 9124/6

xbar = 1520.67

For median:

We arrange first

1508, 1511, 1515, 1528, 1530, 1532

The two values at the centre are 1515 and 1528.

Median = 1515+1528/2

Median = 1521.5

c) To know the measure of central tendency that was most affected, we will look at the difference in the values gotten for both mean and median.

∆Mean = 1520.67-1517.33

∆Mean = 3.34

∆Median = 1521.5 - 1519.5

∆Median = 2.0

It can be seen that the measure of central tendency with greater deviation is the mean. Therefore, the mean is more affected by the data entry error.

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