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Practice each question, then open the answer or worked solution.
The class mark of the class 90-120 is:
(A) 90
(B) 105
(C) 115
(D) 120
(B) 105
Explanation:
Class mark = (upper limit + lower limit) / 2
Then,
Class mark = 120 + 90 / 2
= 210 / 2
= 105.
Hence, option (B) is the correct answer.
The range of the data:
25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 is
(A) 10
(B) 15
(C) 18
(D) 26
(D) 26
Explanation:
According to the question,
The minimum and maximum values of given data are 6 and 32 respectively.
Hence,
Range of the data = 32 – 6
= 26
Hence, option (D) is the correct answer.
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is:
(A) 6
(B) 7
(C) 8
(D) 12
(B) 7
Explanation:
According to the question,
Mid value = 10 cm
Width = 6
We know that,
Lower Limit = mid value of class – (width/2)
= 10 – (6/2)
= 10 – 3
= 7
Hence, option (B) is the correct answer.
The width of each of five continuous classes in a frequency distribution is 5 and the lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
(A) 15
(B) 25
(C) 35
(D) 40
(C) 35
Explanation:
According to the question,
Since width of each five consecutive classes = 5
And, the lower limit of lowest class = 10
We get the data as follows,
10-15
15-20
20-25
25-30
30-35
Hence, we get upper limit of highest class = 35
Hence, option (C) is the correct answer.
Let m be the mid-point and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
(A) 2m + l
(B) 2m – l
(C) m – l
(D) m – 2l
(B) 2m – l
Explanation:
Let x be the lower class limit of a continuous frequency distribution
Let y be the upper class limit of a continuous frequency distribution
Now, according to the question,
Mid-point of a class = (x+y) /2
=m
x+y = 2m or, x+l=2m
It is given that, y = l = upper class limit
x=2m-l
Hence, the lower class limit of the class = 2m-l
Hence, option (B) is the correct answer.
The class marks of a frequency distribution are given as follows:
15, 20, 25, …
The class corresponding to the class mark 20 is:
(A) 12.5 – 17.5
(B) 17.5 – 22.5
(C) 18.5 – 21.5
(D) 19.5 – 20.5
(B) 17.5 – 22.5
Explanation:
According to the question,
Difference between class marks = 5
The classes for these class marks are as follows:
2.5 – 7.5
7.5 – 12.5
12.5 – 17.5
17.5 – 22.5
22.5 – 27.5
Hence, the class corresponding to class mark 20 = 17.5 – 22.5
Hence, option (B) is the correct answer.
In the class intervals 10-20, 20-30, the number 20 is included in:
(A) 10-20
(B) 20-30
(C) both the intervals
(D) none of these intervals
(B) 20-30
Explanation:
20 belongs to the group 20-30.
Hence, option (B) is the correct answer.
A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data :
268, 220, 368, 258, 242, 310, 272, 342,
310, 290, 300, 320, 319, 304, 402, 318,
406, 292, 354, 278, 210, 240, 330, 316,
406, 215, 258, 236.
The frequency of the class 310-330 is:
(A) 4
(B) 5
(C) 6
(D) 7
(C) 6
Explanation:
Arranging the following data in a table, we get,
| Class Interval | Frequency |
| 210 – 230 | 3 |
| 230 – 250 | 3 |
| 250 – 270 | 3 |
| 270 – 290 | 2 |
| 290 – 310 | 4 |
| 310 – 330 | 6 |
| 330 – 350 | 2 |
| 350 – 370 | 2 |
| 370 – 390 | 0 |
| 390 – 410 | 3 |
Therefore, we get from the above table,
The frequency of the class 310-330 is = 6
Hence, option (C) is the correct answer.
A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88,
40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96,
102, 110, 88, 74, 112, 14, 34, 44.
The number of classes in the distribution will be :
(A) 9
(B) 10
(C) 11
(D) 12
(B) 10
Explanation:
According to the question,
The given frequency varies from 14 to 112.
So the class intervals are as follows:
13-22, 23-32, 33-42, 43-52, 53-62, 63-72, 73-82, 83-92, 93-102, 103-112.
Number of class interval = 10.
Hence, option (B) is the correct answer.
To draw a histogram to represent the following frequency distribution:
the adjusted frequency for the class 25-45 is :
(A) 6
(B) 5
(C) 3
(D) 2
(D) 2
Explanation:
Adjusted frequency of a class is given by,
(Minimum class size of frequency distribution × Frequency of given class) / Class size of given class
Therefore,
Adjusted frequency for the class 25 – 45 = (5 × 8) / 20 = 2
Hence, option (D) is the correct answer.