The monthly rents (in dollars) paid by 8 people are given below. (Note that these are already ordered from least to greatest.) 545, 945, 970, 995, 1015, 1030, 1040, 1124 Suppose that one of the people moves. His rent changes from $545 to $905. Answer the following. (a) What happens to the median? It decreases by s[ It increases by se It stays the same. It decreases bys] (b) What happens to the mean? It increases by It stays the same.

Step-by-step explanation:

Before: 545, 945, 970, 995 , 1015, 1030 , 1040 , 1124

Now:  905, 945, 970, 995 , 1015, 1030 , 1040 , 1124

a) Median stays same.

As number of data's is 8, median is the average of 4th and 5th term

Median = (995 + 1015)/2 = 2010/2  =

Median = 1005

When rent changes from $545 to $ 905, the 4th and 5th term remains same.

So, median stays same

b) Mean increases 45

Before: 545, 945, 970, 995 , 1015, 1030 , 1040 , 1124

Mean = Sum of rents ÷ 8

= (545 + 945+ 970+ 995 + 1015 +1030 + 1040 + 1124)/8

=7664/8

= 958

Now : 905, 945, 970, 995 , 1015, 1030 , 1040 , 1124

Mean = (905 + 945+ 970+ 995 + 1015 +1030 + 1040 + 1124)/8

         = 8024/8

         = 1003

Mean increases by 1003 - 958 = 45

No, Micah is not correct. Since there are equal numbers of people in each group who drive and who do not drive, the relative frequency for each is 50%. Because the relative frequencies are the same, there is no association between the variables.