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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?

User Lou K
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Answer:

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.

User Minisaurus
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