
Concept explainers
1.
Find the
1.

Answer to Problem 11CAP
The range of data in the populationis 74.
Explanation of Solution
Calculation:
The data measures the number of minutes (per day)that a small hypothetical population of college students spendsonline.
Range:
The difference between the largest value in the data set and smallest value in the data set is termed as range.
The formula for range is,
The largest value is 98 and the smallest value is 24. The range is,
Hence, the value of range is 74.
2.
Find the IQR of data in this population.
2.

Answer to Problem 11CAP
The IQR of data in this population is 17.
Explanation of Solution
Calculation:
Arrange the data in the numeric order 24, 65, 77, 82, 88, 88, 92, 94, 98, 98. That is
The formula to find the
Follow the below steps to calculate IQR for given data:
The position of
First
The median of the lower half of the values from
Compute position of
The position of first quartile for data is,
The value that is in position 3 is 77. The value of lower quartile
Third quartile:
The median of the upper half of the values from
Compute position of
The position of third quartile for data is,
The value that is in position 3 is 94. The value of upper quartile
The interquartile range is,
Hence, the IQR of data in this population is 17.
3.
Find the SIQR of data in this population.
3.

Answer to Problem 11CAP
The SIQR of data in this population is 8.5.
Explanation of Solution
Calculation:
The formula to find the semi-interquartile range is,
From part 2, the IQR value is 17.
The semi-interquartile range is,
Hence, the SIQR of data in this population is 8.5.
4.
Find the population variance.
4.

Answer to Problem 11CAP
The population variance is 448.64.
Explanation of Solution
Calculation:
The formula for the sum of the squared deviation is,
Assume that, the random variable X denotes the scores of students.
The formula for population variance is,
Follow the below steps to find the sum of the squared deviation for given data.
The population meanis,
The squared deviation for score 98 is,
Similarly, the squared deviation for the remaining scores is obtained as shown in table (1).
Scores | Squared deviation |
98 | |
77 | |
88 | |
65 | |
24 | |
92 | |
94 | |
98 | |
88 | |
82 |
Table 1
The sum of squared deviation is,
The population varianceis,
Hence, the population varianceis 448.64.
5.
Find the population standard deviation.
5.

Answer to Problem 11CAP
The population standard deviationis 21.18.
Explanation of Solution
Calculation:
The population variance is 448.64.
The formula for the standard deviation is,
The standard deviation is,
Hence, the population standard deviation is 21.18.
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Chapter 4 Solutions
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