Questions: During a quality assurance check, the actual contents (in grams) of six containers of protein powder were recorded as 1523, 1529, 1500, 1517, 1531, and 1510. (a) Find the mean and the median of the contents. (b) The third value was incorrectly measured and is actually 1521. 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?

Solution Steps

The original data set is \([1523, 1529, 1500, 1517, 1531, 1510]\).

• Mean: The mean is calculated by summing all the values and dividing by the number of values: \[ \text{Mean} = \frac{1523 + 1529 + 1500 + 1517 + 1531 + 1510}{6} = \frac{9110}{6} \approx 1518.33 \]

• Median: To find the median, first arrange the data in ascending order: \([1500, 1510, 1517, 1523, 1529, 1531]\). Since there is an even number of data points, the median is the average of the two middle values: \[ \text{Median} = \frac{1517 + 1523}{2} = \frac{3040}{2} = 1520 \]

The corrected data set is \([1523, 1529, 1521, 1517, 1531, 1510]\).

• Mean: The mean is calculated similarly: \[ \text{Mean} = \frac{1523 + 1529 + 1521 + 1517 + 1531 + 1510}{6} = \frac{9131}{6} \approx 1521.83 \]

• Median: Arrange the corrected data in ascending order: \([1510, 1517, 1521, 1523, 1529, 1531]\). The median is the average of the two middle values: \[ \text{Median} = \frac{1521 + 1523}{2} = \frac{3044}{2} = 1522 \]

• Mean Difference: The difference in the mean due to the correction is: \[ \text{Mean Difference} = 1521.83 - 1518.33 = 3.5 \]

• Median Difference: The difference in the median due to the correction is: \[ \text{Median Difference} = 1522 - 1520 = 2 \]

Since the mean difference is greater than the median difference, the mean was affected more by the data entry error.

Final Answer