Question

the data - set contains six values: 10, 12, 12, 16, 14, 16. which of the following is an accurate statement when 16 is added to the data - set? a. the median does not change. b. the mean decreases. c. the mean does not change.

Explanation:

Step1: Calculate original mean

The original data - set is \(10,12,12,14,16,16\). The mean \(\bar{x}_1=\frac{10 + 12+12+14+16+16}{6}=\frac{80}{6}=\frac{40}{3}\approx13.33\).

Step2: Calculate original median

Arrange the original data - set in ascending order: \(10,12,12,14,16,16\). Since \(n = 6\) (even), the median \(M_1=\frac{12 + 14}{2}=13\).

Step3: Add 16 to the data - set and calculate new mean

The new data - set is \(10,12,12,14,16,16,16\). The new mean \(\bar{x}_2=\frac{10 + 12+12+14+16+16+16}{7}=\frac{96}{7}\approx13.71\). Since \(\frac{96}{7}>\frac{40}{3}\), the mean increases.

Step4: Add 16 to the data - set and calculate new median

Arrange the new data - set in ascending order: \(10,12,12,14,16,16,16\). Since \(n = 7\) (odd), the median \(M_2 = 14\).

Since the original median is 13 and the new median is 14, the median changes. Since the mean increases from \(\frac{40}{3}\) to \(\frac{96}{7}\), options B and C are incorrect.

Answer:

None of the above options are correct.