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Algebra Statistics Worksheet
Practice Problems
From the data set, the maximum value is 77 and the minimum value is 60.
So, the range of the scores is 77 - 60 = 17.
36 is the outlier of the data as it is far apart from other data values.
So, it may skew the central tendency.
As the outlier influences the mean, the median is the better measure of the central tendency.
Mean of the data set =s u m o f t h e d a t a v a l u e s n u m b e r o f d a t a v a l u e s
[Formula.]
=5+9+11+6+7+6+8+9+11 9
[Substitute the values.]
=72 9
[Add the data values in the numerator.]
= 8
[Divide.]
The data set in the ascending order is 5, 6, 6, 7, 8, 9, 9, 11 and 11.
Median of the set is 8.
[Median is the middle data value of the ordered set.]
Mode is/are the data value(s) which appear most often in the data set.
In the set, the values 6, 9 and 11 appeared two times and all others appeared only once.
So, the modes of the data set are 6, 9 and 11.
The ordered data set is 57, 58, 59, 61, 62, 63, 64, 64, 64 and 74.
As there are 10 data values in the set of data, the median of the data set is the average of the two middle numbers.
So, the median of the data set =62+63 2 125 2
In the graph, 1 is much lower than the other data values.
So, 1% is the outlier, which is the loss percentage in the month of May.
= 100 - 90 = 10
IQR = upper quartile - lower quartile
[upper quartile = 100 and lower quartile = 90.]
Lower quartile - (1.5 × IQR) = 90 - (1.5 × 10) = 75
[Substitute and subtract.]
Upper quartile + (1.5 × IQR) = 100 + (1.5 × 10) = 115
[Substitute and simplify.]
In the box and whisker plot, all the values are between 82 and 106. So, there is no value less than 75 or greater than 115.
So, there is no outlier.
Sum of the data values of the data set = 14 + 20 + 8 + 4 + 6 + 12 + 16 + 22 + 36 + 13 = 151
Mean =151 10
So, the mean of the data set is 15.1.
The ascending order of the above data set is: 25, 25, 30, 35, 40, 40, 40, 50, 50, 75.
In the above list of data, least value is 25 and the greatest value is 75.
Middle quartile = median of the data =(40+40) 2
Lower quartile = median of lower half of the data = 30
Upper quartile = median of upper half of the data = 50
So, among the plots, Plot 3 is the equivalent box-and-whisker plot of the data.
Algebra Statistics Worksheet
- Page 1
1.
The scores of 10 students in Mr. Henry's class are 65, 60, 64, 61, 66, 67, 67, 67, 62 and 77. Find the range of the scores.
| a. | 19 | ||
| b. | 16 | ||
 | c. | 17 | |
| d. | 18 |
Solution:
Range of a data set is the difference between the maximum and minimum data values of the data set.From the data set, the maximum value is 77 and the minimum value is 60.
So, the range of the scores is 77 - 60 = 17.
Correct answer : (3)
2.
The graph of a normal distribution will be ________.
| a. | It has no predictable shape | ||
| b. | Skewed left | ||
| c. | Skewed right | ||
| d. | Symmetric |
Solution:
The graph of a normal distribution will be symmetric.Correct answer : (4)
3.
Mean Deviation of a data value is ______ .
| a. | how much it differs from the mode. | ||
| b. | sum of all the data values. | ||
| c. | range of the given data set. | ||
| d. | how much it differs from the mean. |
Solution:
Mean Deviation of a data value is how much it differs from the mean.Correct answer : (4)
4.
Dennis has taken an aptitude test 8 times and his scores are 96, 98, 98, 105, 36, 87, 95 and 93. Is mean or median the better measure of central tendency?
| a. | Mean | ||
| b. | Median |
Solution:
Mean is a better measure of central tendency if there is no outlier for the data.36 is the outlier of the data as it is far apart from other data values.
So, it may skew the central tendency.
As the outlier influences the mean, the median is the better measure of the central tendency.
Correct answer : (2)
5.
Find the measures of central tendency for the data set 5, 9, 11, 6, 7, 6, 8, 9 and 11.
| a. | Mean = 8, median = 8 and modes are 6 | ||
| b. | Mean = 8, median = 11 and modes are 6, 9 and 11 | ||
| c. | Mean = 8, median = 8 and modes are 6 and 11 | ||
| d. | Mean = 8, median = 8 and modes are 6, 9 and 11 |
Solution:
Mean, median and mode of a data set are the measures of central tendency.Mean of the data set =
[Formula.]
=
[Substitute the values.]
=
[Add the data values in the numerator.]
= 8
[Divide.]
The data set in the ascending order is 5, 6, 6, 7, 8, 9, 9, 11 and 11.
Median of the set is 8.
[Median is the middle data value of the ordered set.]
Mode is/are the data value(s) which appear most often in the data set.
In the set, the values 6, 9 and 11 appeared two times and all others appeared only once.
So, the modes of the data set are 6, 9 and 11.
Correct answer : (4)
6.
Find the median for the data set which shows the scores of 10 students in Mr. Chris's class 62, 57, 61, 58, 63, 64, 64, 64, 59 and 74.
| a. | 63 | ||
| b. | 62.5 | |
| c. | 64 | ||
| d. | 62 |
Solution:
Median is the middle data value of an ordered data set.The ordered data set is 57, 58, 59, 61, 62, 63, 64, 64, 64 and 74.
As there are 10 data values in the set of data, the median of the data set is the average of the two middle numbers.
So, the median of the data set =
Correct answer : (2)
7.
The loss percentages of a company for 6 months are given in a bar graph. What is the outlier in the data?
| a. | 1% | |
| b. | 5% | ||
| c. | 6% | ||
| d. | 4% |
Solution:
Outlier is the value that is much higher or lower than the other data values.In the graph, 1 is much lower than the other data values.
So, 1% is the outlier, which is the loss percentage in the month of May.
Correct answer : (1)
8.
What is the outlier in the box and whisker plot?
| a. | 92 | ||
| b. | 84 | ||
| c. | 90 | ||
| d. | No outlier |
Solution:
In a box and whisker plot, outlier is the data value that should be less than [lower quartile - (1.5 × IQR)] or greater than [upper quartile + (1.5 × IQR)].= 100 - 90 = 10
IQR = upper quartile - lower quartile
[upper quartile = 100 and lower quartile = 90.]
Lower quartile - (1.5 × IQR) = 90 - (1.5 × 10) = 75
[Substitute and subtract.]
Upper quartile + (1.5 × IQR) = 100 + (1.5 × 10) = 115
[Substitute and simplify.]
In the box and whisker plot, all the values are between 82 and 106. So, there is no value less than 75 or greater than 115.
So, there is no outlier.
Correct answer : (4)
9.
Find the mean of the data set: 14, 20, 8, 4, 6, 12, 16, 22, 36, 13
| a. | 17.1 | ||
| b. | 14.1 | ||
| c. | 16.1 | ||
| d. | 15.1 |
Solution:
Mean is the sum of the data values divided by the number of data items in a data set.Sum of the data values of the data set = 14 + 20 + 8 + 4 + 6 + 12 + 16 + 22 + 36 + 13 = 151
Mean =
So, the mean of the data set is 15.1.
Correct answer : (4)
10.
The number of games won by a famous basketball team between 1991 and 2000 are 25, 30, 25, 50, 40, 75, 40, 50, 35 and 40. Which of the box-and-whisker plots represents the data?
| a. | Plot 1 | ||
| b. | Plot 2 | ||
| c. | Plot 3 | |
| d. | Plot 4 |
Solution:
The number of games won by the team each year from the year 1991 to the year 2000 are: 25, 30, 25, 50, 40, 75, 40, 50, 35 and 40.The ascending order of the above data set is: 25, 25, 30, 35, 40, 40, 40, 50, 50, 75.
In the above list of data, least value is 25 and the greatest value is 75.
Middle quartile = median of the data =
Lower quartile = median of lower half of the data = 30
Upper quartile = median of upper half of the data = 50
So, among the plots, Plot 3 is the equivalent box-and-whisker plot of the data.
Correct answer : (3)