Mathematics
Directions:
Study the following table carefully and answer the questions that follow:
| Age (in years) | Number of employees (Frequency) | Cumulative frequency |
|---|---|---|
| 30 - 35 | 5 | 5 |
| 35 - 40 | 7 | 12 |
| 40 - 45 | 6 | 18 |
| 45 - 50 | 9 | 27 |
| 50 - 55 | 4 | 31 |
Based on above table, answer the following questions:
- The total number of employees is :
(a) 30
(b) 31
(c) 55
(d) Cannot be determined
- How many employees are less than 50 years of age?
(a) 9
(b) 18
(c) 27
(d) 31
- How many employees are atleast 40 years old?
(a) 12
(b) 18
(c) 19
(d) 27
- What is the difference between the class-mark of the first and last class intervals?
(a) 20
(b) 22.5
(c) 25
(d) 27.5
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Answer
17. The total number of employees in a frequency distribution is the sum of all individual frequencies, which is equal to the cumulative frequency of the last class interval.
∴ The total number of employees is 31.
Hence, Option (b) is the correct option.
18. As we know,
In a grouped data of exclusive form, the data related to upper limit is excluded.
Given, frequency distribution are 30 - 35, 35 - 40, 40 - 45, 45 - 50.
Since, number of employees less than 50 years of age corresponds to the cumulative frequency for the class 45 - 50 = 27.
Hence, Option (c) is the correct option.
19. Given, atleast 40 years old,
The classes includes, 40 - 45, 45 - 50 and 50 - 55.
Frequency of classes = 6 + 9 + 4 = 19
Hence, Option (c) is the correct option.
20. We know that,
Class mark =
Class mark of first class = = 32.5
Class mark of last class = = 52.5
Difference = 52.5 - 32.5 = 20
Hence, Option (a) is the correct option.
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Related Questions
If the lower limit of a class-interval is 48 and the class-mark is 55, then the upper limit is :
60
62
64
61
If the class-intervals in a frequency distribution are 1 - 11, 11 - 21, 21 - 31, etc., then class 1 - 11 means :
more than 1 and less than 11
1 or more but less than 11
equal to or more than 1 but equal to or less than 11
more than 1 but less than or equal to 11
Assertion (A) : For constructing frequency polygon, class-marks should be calculated.
Reason (R) : To construct a frequency polygon, we take class marks along x-axis and corresponding frequencies along y-axis.
A is true, R is the false
A is false, R is true
Both A and R are true
Both A and R are false.
Assertion (A) : If the class-mark of a class is 9.5 and the class size is 6, then the class interval is 6 - 12.
Reason (R) : Class mark =
A is true, R is the false
A is false, R is true
Both A and R are true
Both A and R are false.