4. an accountant calculated the total salaries paid by each client corporation.
corporation total yearly compensation
a $55,908,440
b $25,748,440
c $18,174,180
d $18,174,180
e $4,881,840
a. find the mean.
b. find the median.
c. find the mode.
d. find the range.
e. do there seem to be one or more outliers? if so, what numbers are outliers? do you think the outliers skew the mean?
it is often helpful to calculate measures of central tendency and variation without outliers so that the values are not skewed.

Question

4. an accountant calculated the total salaries paid by each client corporation.
corporation  total yearly compensation
a  $55,908,440
b  $25,748,440
c  $18,174,180
d  $18,174,180
e  $4,881,840
a. find the mean.
b. find the median.
c. find the mode.
d. find the range.
e. do there seem to be one or more outliers? if so, what numbers are outliers? do you think the outliers skew the mean?
it is often helpful to calculate measures of central tendency and variation without outliers so that the values are not skewed.

Accepted Answer

### Part A: Find the Mean
# Explanation:
## Step 1: Sum the values
Sum the total yearly compensations: $ 55908440 + 25748440 + 18174180 + 18174180 + 4881840 $
Calculate the sum: $ 55908440+25748440 = 81656880 $; $ 81656880+18174180 = 99831060 $; $ 99831060+18174180 = 118005240 $; $ 118005240 + 4881840 = 122887080 $
## Step 2: Divide by the number of values
There are 5 corporations, so mean $ = \frac{122887080}{5} $
Calculate the mean: $ \frac{122887080}{5} = 24577416 $
# Answer:
The mean is $\$24,577,416$

### Part B: Find the Median
# Explanation:
## Step 1: Order the data
Order the total yearly compensations from least to greatest: $ 4881840, 18174180, 18174180, 25748440, 55908440 $
## Step 2: Find the middle value
Since there are 5 values (odd number), the median is the 3rd value.
# Answer:
The median is $\$18,174,180$

### Part C: Find the Mode
# Explanation:
## Step 1: Identify the most frequent value
Check the frequency of each value: $ 4881840 $ appears once, $ 18174180 $ appears twice, $ 25748440 $ appears once, $ 55908440 $ appears once.
The value $ 18174180 $ appears most frequently (twice).
# Answer:
The mode is $\$18,174,180$

### Part D: Find the Range
# Explanation:
## Step 1: Identify maximum and minimum
Maximum value: $ 55908440 $, Minimum value: $ 4881840 $
## Step 2: Subtract minimum from maximum
Range $ = 55908440 - 4881840 $
Calculate the range: $ 55908440 - 4881840 = 51026600 $
# Answer:
The range is $\$51,026,600$

### Part E: Outliers and Skewness
# Explanation:
## Step 1: Check for outliers (using IQR method)
First, find Q1 and Q3. The data is ordered: $ 4881840, 18174180, 18174180, 25748440, 55908440 $
Q2 (median) is $ 18174180 $. The lower half is $ [4881840, 18174180] $, Q1 is the median of the lower half: $ \frac{4881840 + 18174180}{2} = \frac{23056020}{2} = 11528010 $
The upper half is $ [18174180, 55908440] $, Q3 is the median of the upper half: $ \frac{18174180 + 25748440}{2} = \frac{43922620}{2} = 21961310 $
## Step 2: Calculate IQR
IQR $ = Q3 - Q1 = 21961310 - 11528010 = 10433300 $
## Step 3: Check for outliers
Lower bound: $ Q1 - 1.5 \times IQR = 11528010 - 1.5 \times 10433300 = 11528010 - 15649950 = -4121940 $ (no lower outliers as values are positive)
Upper bound: $ Q3 + 1.5 \times IQR = 21961310 + 1.5 \times 10433300 = 21961310 + 15649950 = 37611260 $
Values above 37611260: $ 55908440 $ is above, and $ 25748440 $ is below. Also, $ 4881840 $ is far from the other lower values? Wait, re - evaluate. Alternatively, by visual inspection, $ 55908440 $ and $ 4881840 $ are far from the cluster of $ 18174180, 18174180, 25748440 $. But using IQR, $ 55908440>37611260 $, so $ 55908440 $ is an upper outlier. $ 4881840 $: check if it's a lower outlier. Lower bound is negative, so $ 4881840 $ is not a lower outlier (since all values are above the lower bound). Wait, maybe another approach: the values $ 4881840 $ and $ 55908440 $ are far from the middle values. But from IQR, only $ 55908440 $ is an upper outlier. However, $ 4881840 $ is much smaller than the other four values (the other four are around 18 - 55 million, and $ 4.88 $ million is much smaller). So possible outliers: $ \$4,881,840 $ and $ \$55,908,440 $
## Step 4: Skewness of mean
The mean is $ 24577416 $. The median is $ 18174180 $. The mean is greater than the median, which suggests right - skewness. The large outlier ($ 55908440 $) and the small outlier ($ 4881840 $): the large outlier pulls the mean up, and the small outlier pulls it down, but the large outlier has a bigger impact.
# Answer:
- Outliers: Yes, $\$4,881,840$ and $\$55,908,440$ seem to be outliers (or at least $\$55,908,440$ is an outlier by IQR, and $\$4,881,840$ is much smaller than the rest).
- Skewness: The outlier $\$55,908,440$ (a large value) skews the mean to be higher than the median. The outlier $\$4,881,840$ (a small value) would skew the mean lower, but the large outlier has a more significant impact here, resulting in the mean being greater than the median.

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QUESTION IMAGE

Question

Explanation:

Part A: Find the Mean

Step 1: Sum the values

Step 2: Divide by the number of values

Step 1: Order the data

Step 2: Find the middle value

Step 1: Identify the most frequent value

Answer:

Part B: Find the Median