Question 1201976
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Part (a)


Spreadsheet software is strongly recommended for this problem.
I'm using LibreOffice Calc spreadsheet.


The sorted data looks like this
45,45,50,60,70
75,75,75,75,75
75,75,75,75,75
75,75,75,75,80
80,100,100,100,100
100,100,100,100,100
100,100,100,100,100
100,100,100,100,105
120,120,130,150,150
150,150,150,150,150
150,150,150,150,150
150,160,160,165,170
170,170,200,200,200
200,200,200,200,220
250,250,300,300,320
400,400,400,450,500

The block of numbers has 16 rows and 5 columns
16*5 = 80 values total


min = 45
max = 500
range = max - min = 500-45 = 455


Divide this into 10 groups
455/10 = 45.5


Each group is about 46 units wide.


The class intervals are:
<table border = "1" cellpadding = "5"><tr><td>Number</td><td>Class Interval</td></tr><tr><td>1</td><td>45 - 91</td></tr><tr><td>2</td><td>92 - 138</td></tr><tr><td>3</td><td>139 - 185</td></tr><tr><td>4</td><td>186 - 232</td></tr><tr><td>5</td><td>233 - 279</td></tr><tr><td>6</td><td>280 - 326</td></tr><tr><td>7</td><td>327 - 373</td></tr><tr><td>8</td><td>374 - 420</td></tr><tr><td>9</td><td>421 - 467</td></tr><tr><td>10</td><td>468 - 514</td></tr></table>


There are 21 values between 45 and 91
Those values are: 
45,45,50,60,70
75,75,75,75,75
75,75,75,75,75
75,75,75,75,80
80


Repeat the same idea for the other class intervals to get this grouped frequency distribution.
<table border = "1" cellpadding = "5"><tr><td>Class Interval</td><td>Frequency</td></tr><tr><td>45 - 91</td><td>21</td></tr><tr><td>92 - 138</td><td>22</td></tr><tr><td>139 - 185</td><td>19</td></tr><tr><td>186 - 232</td><td>8</td></tr><tr><td>233 - 279</td><td>2</td></tr><tr><td>280 - 326</td><td>3</td></tr><tr><td>327 - 373</td><td>0</td></tr><tr><td>374 - 420</td><td>3</td></tr><tr><td>421 - 467</td><td>1</td></tr><tr><td>468 - 514</td><td>1</td></tr></table>


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Part (b)


Use spreadsheet software to create the histogram. 
This is what it looks like
*[illustration Screenshot_166.png]
Each bar's height is due to the frequency for that specific class interval.
Example: There are 21 people in the first class interval, which means the left-most bar is 21 units tall.



I'll let you do the frequency polygon.


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Part (c)


Each number mentioned is a positive whole number. 
This data is discrete.


Midpoints between some values make sense (eg: between 100 and 200), but the midpoint between something like 105.11 and 105.12 does not make sense because any currency has a smallest atomic element.


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Part (d)


To find the mean, add up the values and divide by n = 80 because that's how many values there are in the set.


Use the spreadsheet software to quickly add the values up (use the command called "SUM")
The values add 11890


Then 11890/n = 11890/80 = 148.625 is the arithmetic mean.


The median is found between slot 40 and 41 (because n/2 = 80/2 = 40)
The values in these slots are 105 and 120. The midpoint of them is (105+120)/2 = 112.5


mean = 148.625
median = 112.5


The mean and median represent the center of the data distribution. 
Since mean > median in this case, the data is skewed to the right (i.e. positively skewed)


Right-skewed data occurs when you have large outlier(s).


Notice that the histogram (in part b) has a tail stretched to the right.


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Part (e)


Use the spreadsheet software to compute the variance. 
Doing so by hand would be tedious busy-work.


sample variance = 8887.32594936709 approximately


sample standard deviation = sqrt(sample variance)
sample standard deviation = sqrt(8887.32594936709)
sample standard deviation = 94.2726150553123



Summary:
sample variance = 8887.32594936709 
sample standard deviation = 94.2726150553123
Each value is approximate

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