Question

during a quality assurance check, the actual contents (in grams) of six containers of protein powder were recorded as 1524, 1526, 1504, 1517, 1536, and 1513. (a) find the mean and the median of the contents. (b) the third value was incorrectly measured and is actually 1519. 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? (a) the mean is . (round to one decimal place as needed.)

Explanation:

Step1: Calculate sum for original mean

Sum of original data: $1524 + 1526+1504 + 1517+1536+1513=9120$

Step2: Calculate original mean

Mean = $\frac{9120}{6}=1520.0$

Step3: Arrange original data in ascending order

$1504, 1513, 1517, 1524, 1526, 1536$

Step4: Calculate original median

Since $n = 6$ (even), median = $\frac{1517 + 1524}{2}=1520.5$

Step5: Correct data and calculate new sum

New sum with corrected value: $1524+1526 + 1519+1517+1536+1513=9135$

Step6: Calculate new mean

New mean = $\frac{9135}{6}=1522.5$

Step7: Arrange new - data in ascending order

$1513, 1517, 1519, 1524, 1526, 1536$

Step8: Calculate new median

Since $n = 6$ (even), new median = $\frac{1519 + 1524}{2}=1521.5$

Step9: Calculate change in mean and median

Change in mean: $1522.5-1520.0 = 2.5$
Change in median: $1521.5 - 1520.5=1.0$

Answer:

(a) Mean: $1520.0$
(b) Mean: $1522.5$, Median: $1521.5$
(c) The mean was affected more by the data - entry error.