I need help! Please help asap this was supposed to be due yesterday! Effects of Shifting, Adding, and Removing Data

3. Kay owns a small business that employees 6 other people. Kay makes $90,000 per year, and the other 6 employees make between $45,000 and $60,000 per year. Kay decided to increase her salary by $20,000 and leave the rest of the salaries the same. How will increasing her salary affect the mean and median?

• Both the mean and median will increase .
• The mean will increase, and the median will stay the same.
• The median will increase, and the mean will stay the same.
• The mean will increase, and the median will decrease.

B) The mean will increase and the median will stay the same.

Explanation:

Let's look at a small example of why this works.

Suppose that we had the set of 5 items {1,2,3,4,5}. The value 3 is right in the middle so this is the median. The mean is also 3 because adding all the values gets us 15 and then we divide over 5 (the number of items in the set) to get 15/5 = 3.

For any symmetric set of data like this, the mean and median are the same value.

Now let's change that last item 5 to something larger. Let's bump it up to 10

The new data set is {1,2,3,4,10}. The median is still 3 as it's still in the middle. But now the mean is (1+2+3+4+10)/5 = 20/5 = 4. The old mean was 3, and the new mean is 4. So we have the mean increased thanks to that 5 increasing to 10.

We could increase 10 to 20 getting {1,2,3,4,20}. The median is still the same. The mean is now (1+2+3+4+20)/5 = 30/5 = 6. So we've increased the mean again. This process can be continued on for as long as you like.

The outliers to the right hand side effectively pull on the mean in a similar analogy to how gravity pulls on an object. So this is one way to remember that the mean is effected by outliers. The median only cares about being in the middle so that's why it doesn't change. If we changed the number of items in the data set, then it's a different story. However, Kay is only changing her salary rather than bring on a new hire (or two).

So that small example can be applied to Kay's situation even though the numbers are much larger. The median stays the same, but Kay's (much) higher salary pulls on the mean to make it larger than it should be. This is often why a trimmed mean is used when looking at centers of data sets. For data sets that have very large outliers, the median is the more reliable measure of center.

Step-by-step explanation:

18.2/3.14 = 5.79617834 (diameter)

5.79617834/2 = 2.89808917(radius)

Area = 3.14 * r^2

2.89808917^2 = 8.39892084

3.14 * 8.39892084 = 26.4 or 26