Question

during a quality assurance check, the actual contents (in grams) of six containers of protein powder were recorded as 1527, 1526, 1506, 1515, 1528, and 1516. (a) find the mean and the median of the contents. (b) the third value was incorrectly measured and is actually 1517. 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?

Explanation:

Part (a)

Mean Calculation

Step1: Sum the data values

We have the data: \(1527, 1526, 1506, 1515, 1528, 1516\). The sum \(S\) is calculated as:
\(S = 1527+1526 + 1506+1515+1528+1516\)
\(= (1527 + 1526)+(1506 + 1515)+(1528 + 1516)\)
\(= 3053+3021 + 3044\)
\(= 3053+6065\)
\(= 9118\)

Step2: Calculate the mean

The mean \(\bar{x}\) is the sum divided by the number of data points \(n = 6\). So,
\(\bar{x}=\frac{S}{n}=\frac{9118}{6}\approx1519.67\) (rounded to two decimal places)

Median Calculation

Step1: Order the data

First, we order the data from smallest to largest: \(1506, 1515, 1516, 1526, 1527, 1528\)

Step2: Find the median

Since \(n = 6\) (even), the median is the average of the \(\frac{n}{2}\)-th and \((\frac{n}{2}+ 1)\)-th values. \(\frac{n}{2}=3\) and \(\frac{n}{2}+ 1 = 4\). The 3rd value is \(1516\) and the 4th value is \(1526\). So the median \(M=\frac{1516 + 1526}{2}=\frac{3042}{2}=1521\)

Part (b)

Mean Calculation

Step1: Adjust the sum

The incorrect value was \(1506\) and the correct value is \(1517\). The difference in the sum is \(1517-1506 = 11\). So the new sum \(S'=9118 + 11=9129\)

Step2: Calculate the new mean

The new mean \(\bar{x}'=\frac{S'}{n}=\frac{9129}{6}=1521.5\)

Median Calculation

Answer:

• For the mean: The original mean was approximately \(1519.67\) and the new mean is \(1521.5\). The change in mean is \(1521.5 - 1519.67=1.83\)
• For the median: The original median was \(1521\) and the new median is \(1521.5\). The change in median is \(1521.5 - 1521 = 0.5\)

Since the change in the mean (\(1.83\)) is greater than the change in the median (\(0.5\)), the mean was affected more by the data entry error.

Final Answers

(a)

• Mean: \(\approx\boldsymbol{1519.67}\) grams
• Median: \(\boldsymbol{1521}\) grams

(b)

• Mean: \(\boldsymbol{1521.5}\) grams
• Median: \(\boldsymbol{1521.5}\) grams

(c)

The \(\boldsymbol{\text{mean}}\) was affected more.