
Concept explainers
i.
To calculate mean, median, mode, range and standard deviation of Rhonda’s earnings
i.

Answer to Problem 19PPS
Mean is 52.95.
Median is 53.
Mode is 53.
Standard deviation is 6.21
Range is 19
Explanation of Solution
Given information:
Rhonda’s Pay | ||
45 | 55 | 53 |
47 | 53 | 54 |
44 | 56 | 59 |
63 | 47 | 53 |
60 | 57 | 62 |
44 | 50 | 45 |
60 | 53 | 49 |
62 | 47 |
Formula used: Mean formula is,
ˉX=∑fXN
Where, f is frequency
N is total number of items
X is item.
Median formula is,
where N cumulative frequency
Mode is the item with the maximum frequency.
Standard deviation is,
SD=√∑(x−ˉx)2n
Where, SD= standard deviation,
x= value
The Range is the difference between the lowest and highest values.
Calculation:
For Mean,
Substituting the values in mean formula,
=45+55+53+47+53+54+44+56+59+63+47+53+60+57+62+44+50+45+60+53+49+62+4723
On solving,
=121823
On solving,
=52.95
Hence, Mean is 52.95
For median,
Sorting the frequency
44,44,45,45,47,47,47,49,50,53,53,53,53,54,55,56,57,59,60,60,62,62,63,
Substituting the values,
=(23+12)thterm
On solving,
=(12)thterm
Substituting the values,
median=53
Median is 12th term from the sorted frequency which is 53.
For mode,
Mode is the item with the maximum frequency. Which is 53
For standard deviation,
Substituting the values in standard deviation formula,
SD=√(45−52.95)2+(55−52.95)2+(53−52.95)2+(47−52.95)2+(53−52.95)2+(54−52.95)2+(44−52.95)2+(56−52.95)2+(59−52.95)2+(63−52.95)2+(47−52.95)2+(53−52.95)2+(60−52.95)2+(57−52.95)2+(62−52.95)2+(44−52.95)2+(50−52.95)2+(45−52.95)2+(60−52.95)2+(53−52.95)2+(49−52.95)2+(62−52.95)2+(47−52.95)223
On solving,
SD=√(−7.95)2+(2.05)2+(0.05)2+(−5.95)2+(0.05)2+(1.05)2+(−8.95)2+(3.05)2+(6.05)2+(10.05)2+(−5.95)2+(0.05)2+(7.05)2+(4.05)2+(9.05)2+(−8.95)2+(−2.95)2+(−7.95)2+(7.05)2+(0.05)2+(−3.95)2+(9.05)2+(−5.95)223
So,
SD=√38.58
On solving,
SD=6.21
Hence, Standard deviation is 6.21
For Range,
Range is the difference between the lowest and highest values.
Range =max−min
=63-44
=19
Hence, range is 19.
ii.
To calculate mean, median, mode, range and standard deviation of Rhonda’s earnings after deducting the $5 bonus.
ii.

Answer to Problem 19PPS
Mean is 47.91.
Median is 48.
Mode is 48.
Standard deviation is 6.21
Range is 19
Explanation of Solution
Given information:
Rhonda’s pay after deducting $5 bonus
Rhonda’s Pay | ||
40 | 50 | 47 |
42 | 48 | 49 |
39 | 51 | 54 |
58 | 42 | 48 |
55 | 52 | 57 |
39 | 45 | 40 |
55 | 48 | 44 |
57 | 42 |
Formula used: Mean formula is,
ˉX=∑fXN
Where, f is frequency
N is total number of items
X is item.
Standard deviation is,
SD=√∑(x−ˉx)2n
Where, SD= standard deviation,
x= value
The Range is the difference between the lowest and highest values.
Calculation:
For Mean,
Substituting the values in mean formula,
=40+50+47+42+48+49+39+51+54+58+42+48+55+52+57+39+45+40+55+48+44+57+4223
On solving,
=47.91
Hence, Mean is 47.91
For median,
Sorting the frequency
39,39,40,40,42,42,42,44,45,48,48,48,48,49,50,51,52,54,55,55,57,57,58
Substituting the values,
=(23+12)thterm
On solving,
=(12)thterm
Substituting the value,
median=48
For mode,
Mode is the item with the maximum frequency which is 48.
For standard deviation,
Substituting the values in standard deviation formula,
SD=√(40-472.91)+(50-47.91)2+(47-47.91)2+(42-47.91)2+(48-47.91)2+(49-47.91)2+(39-49.91)2+(51-47.91)2+(54-47.91)2+(58-47.91)2+(42-47.91)2+(48-47.91)2+(55-47.91)2+(52-47.91)2+(57-47.91)2+(39-47.91)2+(45-47.91)2+(40-47.91)2+(55-47.91)2+(48-47.91)2+(44-47.91)2+(57-47.91)2+(42-47.91)223
On solving,
SD=√(-72.91)+(2.9)2+(-0.91)2+(-5.91)2+(0.09)2+(1.09)2+(-8.91)2+(3.09)2+(6.09)2+(10.09)2+(-5.91)2+(0.09)2+(7.9)2+(4.09)2+(9.09)2+(-8.91)2+(-2.91)2+(-7.91)2+(7.09)2+(0.09)2+(-3.91)2+(9.09)2+(-5.91)223
So,
SD=√38.62
On solving,
SD=6.21
Hence, Standard deviation is 6.21
For Range,
Range is the difference between the lowest and highest values.
=58-39
=19
Hence, range is 19
Chapter 10 Solutions
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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