Question

the numbers of trading cards owned by 10 middle - school students are given below. (note that these are already ordered from least to greatest.) 444, 468, 470, 473, 479, 486, 488, 496, 533, 653 send data to calculator suppose that the number 653 from this list changes to 543. answer the following. (a) what happens to the median? it decreases by . it increases by . it stays the same. (b) what happens to the mean? it decreases by . it increases by . it stays the same.

Explanation:

Step1: Recall median formula for even - numbered data

For a set of $n = 10$ (even) ordered data values $x_1\leq x_2\leq\cdots\leq x_{10}$, the median $M=\frac{x_{\frac{n}{2}}+x_{\frac{n}{2}+1}}{2}$. Here, $x_{\frac{10}{2}} = x_5=479$ and $x_{\frac{10}{2}+1}=x_6 = 486$, so $M_1=\frac{479 + 486}{2}=\frac{965}{2}=482.5$. After changing 653 to 543, the data is still in order and the middle - two values remain the same. So the median stays the same.

Step2: Recall mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Initially, $\sum_{i=1}^{10}x_i=444 + 468+470+473+479+486+488+496+533+653=4990$, and $\bar{x}_1=\frac{4990}{10}=499$. After changing 653 to 543, the new sum is $\sum_{i = 1}^{10}x_i'=4990-653 + 543=4880$, and the new mean $\bar{x}_2=\frac{4880}{10}=488$. The change in the mean is $499 - 488=11$, so the mean decreases by 11.

Answer:

(a) It stays the same.
(b) It decreases by 11.