a random sample of print - o - matic printing companys employee salaries (in dollars) are recorded in the…
a random sample of print - o - matic printing companys employee salaries (in dollars) are recorded in the table below.\n\nsalaries: 57380, 66346, 58456, 65270, 100702, 98837, 64553, 52718, 272500, 73877, 74235, 76029, 90090, 75311, 109739, 108448\n\na) for the data shown above, find the following.\nround the answer in the first blank to 1 decimal place(s).\nin the second blank put the correct unit of measurement.\nfind the mean:\nfind the median:\nfind the range:\nfind the standard deviation:\n\nb) remove the high ceo salary of 272500 dollars from the data.\nrecompute the following.\nfind the mean:\nfind the median:\nfind the range:\nfind the standard deviation:\n\nc) which measure of center was most affected by removing the high ceo salary? select an answer\n\nd) which measure of spread (variation) was most affected by removing the high ceo salary? select an answer\n\nquestion help: video message instructor post to forum
Answer
Explanation:
Step1: Calculate the mean for part a
The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points. $n = 16$, $\sum_{i=1}^{16}x_{i}=57380 + 66346+58456+65270+100702+98837+64553+52718+272500+73877+74235+76029+90090+75311+109739+108448=1514891$ $\bar{x}=\frac{1514891}{16}\approx94680.7$ dollars
Step2: Calculate the median for part a
First, order the data: $52718,57380,58456,64553,65270,66346,73877,74235,75311,76029,90090,98837,100702,108448,109739,272500$ Since $n = 16$ (even), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered data - points. The 8th value is $74235$ and the 9th value is $75311$. Median$=\frac{74235 + 75311}{2}=74773$ dollars
Step3: Calculate the range for part a
The range is the difference between the maximum and minimum values. Range$=272500 - 52718=219782$ dollars
Step4: Calculate the standard deviation for part a
The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$ First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$, sum them up: $\sum_{i = 1}^{16}(x_{i}-\bar{x})^{2}=(57380 - 94680.7)^{2}+(66346 - 94680.7)^{2}+\cdots+(108448 - 94680.7)^{2}=1.67993989\times10^{10}$ $s=\sqrt{\frac{1.67993989\times10^{10}}{15}}\approx105837.7$ dollars
Step5: Calculate the mean for part b
Remove $272500$. Now $n = 15$, $\sum_{i=1}^{15}x_{i}=1514891-272500 = 1242391$ $\bar{x}=\frac{1242391}{15}\approx82826.1$ dollars
Step6: Calculate the median for part b
Order the remaining 15 data - points: $52718,57380,58456,64553,65270,66346,73877,74235,75311,76029,90090,98837,100702,108448,109739$ Since $n = 15$ (odd), the median is the $(\frac{n + 1}{2})$th ordered data - point, which is the 8th value. Median$=74235$ dollars
Step7: Calculate the range for part b
Range$=109739 - 52718=57021$ dollars
Step8: Calculate the standard deviation for part b
First, calculate $(x_{i}-\bar{x})^{2}$ for each of the 15 non - removed $x_{i}$, sum them up: $\sum_{i = 1}^{15}(x_{i}-\bar{x})^{2}=(57380 - 82826.1)^{2}+(66346 - 82826.1)^{2}+\cdots+(109739 - 82826.1)^{2}=2.47979749\times10^{9}$ $s=\sqrt{\frac{2.47979749\times10^{9}}{14}}\approx42191.5$ dollars
Step9: Answer part c
The mean was most affected. The mean changed from approximately $94680.7$ to $82826.1$. The median changed from $74773$ to $74235$.
Step10: Answer part d
The range was most affected. The range changed from $219782$ to $57021$.
Answer:
a) Mean: $94680.7$ dollars, Median: $74773$ dollars, Range: $219782$ dollars, Standard deviation: $105837.7$ dollars b) Mean: $82826.1$ dollars, Median: $74235$ dollars, Range: $57021$ dollars, Standard deviation: $42191.5$ dollars c) Mean d) Range