#### Solved Examples and Worksheet for Comparing Two Related Sets of Data and Expected Value

Q1The test scores of Chris and Mark on grade 5 tests are in order from lowest to highest. Which statement is true about the comparison of the two sets of scores? A. Chris has a higher median than Mark.
B. Chris has a higher mean than Mark.
C. Chris has a lower mean than Mark.
D. Chris has a same median than Mark.

Step: 1
Median is the data value which is the middle one in the ordered data set.
Step: 2
The middle value of Chris′s score is 65. Median of Chris′s score is 65.
Step: 3
The middle value of Mark′s score is 85. Median of Mark′s score is 83.
Step: 4
On comparing, Chris has a lower median than Mark.
[Since 65 is lower than 83.]
Step: 5
The average of the numbers of a data set is called mean. Mean = sum of the data valuesnumber of data values
Step: 6
Mean of Chris′s score is 78 + 68+ 75+ 65 + 80+ 85+ 707 = 74.4
Step: 7
Mean of Mark′s score is 69+ 72+ 80+ 83+ 78+ 80+ 757 = 76.7
Step: 8
On comparing, Chris has a lower mean than Mark.
[Since 74.4 is lower than 76.7.]
Step: 9
So, Chris has a lower mean than Mark.
Correct Answer is :   Chris has a lower mean than Mark.
Q2Which of the following data sets has greater median?SMike
A. Both data sets have equal median
B. Data B
C. Data A

Step: 1
Arrange the data values of Data A in ascending order: 120, 121, 122, 124, 128, 130, 133
Step: 2
Arrange the data values of Data B in ascending order: 112, 119, 120, 125, 126, 127, 128
Step: 3
Median is the middle data value in an ordered data set.
Step: 4
The middle value of Data A is 124. Median of Data A = 124
Step: 5
The middle value of Data B is 125. Median of Data B = 125
Step: 6
On comparing, Data B has a greater median than Data A.
[Since 125 is greater than 124.]
Correct Answer is :   Data B
Q3A cellular company wants to compare the number of calls made by female customers and male customers daily. The company randomly selected 10 female customers and 10 male customers. The cellular company then recorded the number of calls made daily by each customer. Which of the following statement is true?
 Female Customers 10 10 5 32 24 18 12 8 10 11 Male Customers 21 22 15 14 9 10 13 12 14 10

A. The mean of Female customers data is greater than that of Male customers data.
B. The mean of Female customers data is lesser than that of Male customers data.
C. The mean of Female customers data is equal to that of Male customers data.

Step: 1
The average of the numbers of a data set is called mean.
Mean = sum of the data valuesnumber of data values
Step: 2
Mean of the number calls made daily by female customers = 10+10+5+32+24+18+12+8+10+1110 =14010 = 14
[Substitute the values and simplify]
Step: 3
Mean of the number calls made daily by male customers = 21+22+15+14+9+10+13+12+14+1010 =14010 = 14
[Substitute the values and simplify]
Step: 4
On comparing, the mean of the number of calls made daily by female customers is equal to the mean of the number of calls made daily by male customers.
Correct Answer is :   The mean of Female customers data is equal to that of Male customers data.
Q4Given the data sets P and Q as follows, which statement is true about the comparison of the two sets?
 Data P 6.3 6.1 6.7 6.6 6.8 6.4 Data Q 6.3 6.8 6.8 6.3 6.7 6.1

A. Both Data P and Data Q have equal median.
B. Data P has lesser median than Data Q
C. Data P has greater median than Data Q

Step: 1
Arrange the data values of Data P in ascending order: 6.1, 6.3, 6.4, 6.6, 6.7, 6.8
Step: 2
Arrange the data values of Data Q in ascending order: 6.1, 6.3, 6.3, 6.7, 6.8, 6.8
Step: 3
Median is the middle data value in an ordered data set.
Step: 4
Here we have two middle values. When there are two middle values, then Median = sum of the two middle values2
Step: 5
The two middle values of Data P are 6.4 and 6.6. Median of Data P = 6.4+6.62 =132 = 6.5
Step: 6
The two middle values of Data Q are 6.3 and 6.7. Median of Data Q = 6.3+6.72 =132 = 6.5
Step: 7
On comparing, both Data P and Data Q have equal median.
Correct Answer is :   Both Data P and Data Q have equal median.
Q5Which of the following data sets has greater interquartile range?
 Data P 2.4 3.1 2.7 3.6 1.4 2.8 Data Q 3.8 2.5 1.8 1.3 2.4 2.1

A. Data P
B. Data Q
C. Both Data P and Data Q have equal interquartile range

Step: 1
The difference of Upper Quartile and Lower Quartile gives the value of interquartile range.
Step: 2
Arrange the data values of Data P in ascending order and find the lower quartile, upper quartile and interquartile range.
1.4, 2.4, 2.7, 2.8, 3.1, 3.6.
Lower Quartile = Median of lower half of the data = 2.4
Upper Quartile = Median of upper half of the data = 3.1
Interquartile Range = Upper Quartile - Lower Quartile = 3.1 - 2.4 = 0.7
Step: 3
Arrange the data values of Data Q in ascending order and find the lower quartile, upper quartile and interquartile range.
1.3, 1.8, 2.1, 2.4, 2.5, 3.8.
Lower Quartile = Median of lower half of the data = 1.8
Upper Quartile = Median of upper half of the data = 2.5
Interquartile Range = Upper Quartile - Lower Quartile = 2.5 - 1.8 = 0.7
Step: 4
Therefore, Data P and Data Q have equal interquartile range.
Correct Answer is :   Both Data P and Data Q have equal interquartile range
Q6Given the data sets M and N as follows, which statement is true about the comparison of the two sets?
 M 12 18 13 21 17 16 24 15 N 17 11 24 27 15 23 20 13

A. Data M has lesser interquartile range than Data N.
B. Data M has greater interquartile range than Data N.
C. Both Data M and Data N has equal interquartile range.

Step: 1
The difference of Upper Quartile and Lower Quartile gives the value of interquartile range.
Step: 2
Arrange the data values of Data M in ascending order and find the lower quartile, upper quartile and interquartile range.
12, 13, 15, 16, 17, 18, 21, 24.
Lower Quartile = Median of lower half of the data = 13+152 = 14
Upper Quartile = Median of upper half of the data = 18+212 = 19.5
Interquartile Range = Upper Quartile - Lower Quartile = 19.5 - 14 = 5.5
Step: 3
Arrange the data values of Data B in ascending order and find the lower quartile, upper quartile and interquartile range.
11, 13, 15, 17, 20, 23, 24, 27.
Lower Quartile = Median of lower half of the data = 13+152 = 14
Upper Quartile = Median of upper half of the data = 23+242 = 23.5
Interquartile Range = Upper Quartile - Lower Quartile = 23.5 - 14 = 9.5
Step: 4
Therefore, Data M has lesser interquartile range than Data N.
[Since 5.5 is lesser than 9.5]
Correct Answer is :   Data M has lesser interquartile range than Data N.
Q7Rob and Andy put the number of goals scored by them this season as shown below, which one of them has a greater mean number of goals? A. Rob
B. Andy
C. Both have an equal mean number of goals

Step: 1
Mean = sum of the data valuesnumber of data values
[Formula.]
Step: 2
Mean of the given data of Rob's score = 5 + 6 + 6 + 74 = 244 = 6
[Substitute and simplify.]
Step: 3
Mean of the given data of Andy's score = 4 + 6 + 6 + 84 = 244 = 6
[Substitute and simplify.]
Step: 4
On comparing, both have an equal mean number of goals.
Correct Answer is :   Both have an equal mean number of goals
Q8The data represents the percentage of children suffering from vascular diseases in 8 selected areas of 2 different countries. Which statement is true about the comparison of the two sets of data? A. Country 1 has same median as Country 2
B. Country 1 has lesser median than Country 2
C. Country 1 has greater mean than Country 2
D. Country 1 has lesser mean than Country 2

Step: 1
Arrange the data values of Country 1 in ascending order: 1.89%, 2.9%, 5%, 6.83%, 9.01%, 12.91%, 13%, 14.56%.
Step: 2
Arrange the data values of Country 2 in ascending order: 0.09%, 1.02%, 2.9%, 2.9%, 4.86%, 11.23%, 12.91%, 13.02%.
Step: 3
Median is the middle data value in an ordered data set.
Step: 4
Here we have two middle values. When there are two middle values, then Median = sum of the two middle value2
Step: 5
The two middle values of Country 1 are 6.83 and 9.01. Median percentage of Country 1 = (6.83+9.01)2 = 7.92%.
Step: 6
The two middle values of Country 2 are 2.9 and 4.86. Median percentage of Country 2 = (2.9+4.86)2 = 3.88%
Step: 7
On comparing, Country 1 has a greater median than Country 2.
Step: 8
The average of the numbers of a data set is called mean. Mean = sum of the data valuesnumber of data values
Step: 9
Mean percentage of children suffering from vascular diseases in Country 1
= (5 + 6.83 + 2.9 + 14.56 + 9.01 + 1.89 + 12.91 + 13)8
Step: 10
66.18
Step: 11
= 8.263%
Step: 12
Mean percentage of children suffering from vascular diseases in Country 2
= ( 2.9+1.02+12.91+13.02+4.86+2.9+0.09+11.23)8
Step: 13
= 48.938
Step: 14
= 6.116%
Step: 15
On comparing, Country 1 has greater mean percentage than Country 2.
[Since 8.263 is greater than 6.116.]
Step: 16
So, Country 1 has greater mean percentage than country 2.
Correct Answer is :   Country 1 has greater mean than Country 2
Q9Companies A and B each have ten employees. Their hourly pay, in dollars, is shown below.
 Company A 4 5 5 5 5 6 7 6 4 5 Company B 5 5 8 8 8 10 6 5 8 7
If Phillips wants to join the company having higher mean hourly pay, then which company would he join?

A. Company A
B. Company B

Step: 1
The average of the numbers of a data set is called mean.
Mean = sum of the data valuesnumber of data values
Step: 2
Mean hourly pay for Company A = (4+5+5+5+5+6+7+6+4+5)10 =5210 = $5.2 [Substitute the values and simplify] Step: 3 Mean hourly pay for Company B = (5+5+8+8+8+10+6+5+8+7)10 =7010 =$7
[Substitute the values and simplify]
Step: 4
On comparing, Company B has higher mean hourly pay than Company A. So, Phillips would join the Company B.
Correct Answer is :   Company B
Q10The number of stamps of different countries collected by Edward and Richard are given below. Who has the greater standard deviation for this data?
 Edward 90 70 50 45 65 Richard 85 40 60 75 80

A. Both have equal standard deviation
B. Richard
C. Edward

Step: 1
To find the standard deviation first we need to find the mean of the data.
Mean = sum of the data valuesnumber of data values
Step: 2
Mean (X) of stamps of different countries collected by Edward = (90 +70 + 50 + 45 + 65)5 =3205 = 64
Step: 3
Mean (Y) of stamps of different countries collected by Richard = (85 + 40 + 60 + 75 + 80)5 =3405 = 68
Step: 4
Standard deviation = (x1-X)2+(x2-X)2+...+(xn-X)2n
[X = mean of the data, x1, x2, xn represents the data values.]
Step: 5
Standard deviation of stamps collected by Edward = (x1-X)2+(x2-X)2+(x3-X)2+(x4-X)2+(x5-X)2n
= (90-64)2+(70 - 64)2+(50 - 64)2+(45 - 64)2+(65 - 64)25
= 676 + 36 + 196 + 361 + 15
= 12705
= 254
= 15.94
Step: 6
Standard deviation of stamps collected by Richard = (y1-Y)2+(y2-Y)2+(y3-Y)2+(y4-Y)2+(y5-Y)2n
= (85-68)2+(40 - 68)2+(60 - 68)2+(75 - 68)2+(80 - 68)25
= 289 + 784 + 64 + 49 + 1445
= 13305
= 266
= 16.31
Step: 7
On comparing, Richard has greater standard deviation than Edward.