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Practice each question, then open the answer or worked solution.
In the formula
For finding the mean of grouped data di’s are deviations from a of
(A) Lower limits of the classes
(B) Upper limits of the classes
(C) Mid points of the classes
(D) Frequencies of the class marks
(C) Mid points of the classes
Explanation:
We know,
di = xi – a
Where,
xi are data and ‘a’ is the assumed mean
So, di are the deviations from of mid – points of the classes.
Hence, the option (C) is correct
While computing mean of grouped data, we assume that the frequencies are
(A) Evenly distributed over all the classes
(B) Centred at the class marks of the classes
(C) Centred at the upper limits of the classes
(D) Centred at the lower limits of the classes
(B) Centered at the class marks of the classes
Explanation:
In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.
Hence, the option (B) is correct
If xi’s are the mid points of the class intervals of grouped data, fi’s are the corresponding frequencies and x is the mean, then (fixi – \(\begin{array}{l}\overline{x}\end{array} \) ) is equal to
(A)0 (B) –1 (C) 1 (D) 2
(A) 0
Explanation:
Mean (x) = Sum of all the observations/ Number of observations
x = (f1x1 + f2x2 + …..+ fnxn) / f1 + f2 +……+ fn
x = Σfixi / Σfi, Σfi = n
x = Σfixi / n
n x = Σfixi …………… (1)
Σ (fixi – x) = (f1x1 – x) + (f2x2 – x)+ …..+ (fnxn – x)
Σ (fixi – x) = (f1x1 + f2x2 + …..+ fnxn) – (x +x +….n times)
Σ (fixi – x) = Σfixi –nx
Σ(fixi – x) = nx – nx ( From eq1)
Σ(fixi – x) = 0
Hence, option (A) is correct
In the formula x = a + h(fiui/fi), for finding the mean of grouped frequency distribution, ui =
(A) (xi+a)/h
(B) h (xi – a)
(C) (xi –a)/h
(D) (a – xi)/h
(C) (xi –a)/h
Explanation:
According to the question,
x = a + h(fiui/fi),
Above formula is a step deviation formula.
In the above formula,
xi is data values,
a is assumed mean,
h is class size,
When class size is same we simplify the calculations of the mean by computing the coded mean of u1,u2,u3…..,
Where ui = (xi – a)/h
Hence, option (C) is correct
The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(A) mean (B) median
(C) mode (D) all the three above
(B) Median
Explanation:
Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa, the abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
Hence, option (B) is correct
For the following distribution :
Class 0-05 5-10 10-15 15-20 20-25
Frequency 10 15 12 20 9
the sum of lower limits of the median class and modal class is
(A)15 (B) 25 (C) 30 (D) 35
(B) 25
Explanation:
| Class | Frequency | Cumulative Frequency |
| 0-5 | 10 | 10 |
| 5-10 | 15 | 25 |
| 10-15 | 12 | 37 |
| 15-20 | 20 | 57 |
| 20-25 | 9 | 66 |
From the table, N/2 = 66/2 = 33, which lies in the interval 10 – 15.
Hence, lower limit of the median class is 10.
The highest frequency is 20, which lies in between the interval 15 – 20.
Hence, lower limit of modal class is 15.
Therefore, required sum is 10 + 15 = 25.
Hence, option (B) is correct
Consider the following frequency distribution:
Class 0-05 6-11 12-17 18-23 24-29
Frequency 13 10 15 8 11
The upper limit of the median class is
(A)17 (B) 17.5 (C) 18 (D) 18.5
(B) 17.5
Explanation:
According to the question,
Classes are not continuous, hence, we make the data continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.
| Class | Frequency | Cumulative Frequency |
| 0.5-5.5 | 13 | 13 |
| 6.5-11.5 | 10 | 23 |
| 11.5-17.5 | 15 | 38 |
| 17.5-23.5 | 8 | 46 |
| 23.5-29.5 | 11 | 57 |
According to the question,
N/2 = 57/2 = 28.5
28.5 lies in between the interval 11.5 – 17.5.
Therefore, the upper limit is 17.5.
Hence, option (B) is correct
For the following distribution:
Marks Number of students
Below 10 3
Below 20 12
Below 30 27
Below 40 57
Below 50 75
Below 60 80
The modal class is
(A)10-20 (B) 20-30 (C) 30-40 (D) 50-60
(C) 30-40
Explanation:
| Marks | Number of students | Cumulative Frequency |
| Below 10 | 3=3 | 3 |
| 10-20 | (12 – 3) = 9 | 12 |
| 20-30 | (27 – 12) = 15 | 27 |
| 30-40 | (57 – 27) = 30 | 57 |
| 40-50 | (75 – 57) = 18 | 75 |
| 50-60 | (80 – 75) = 5 | 80 |
Here, we see that the highest frequency is 30, which lies in the interval 30 – 40.
Hence, option (C) is correct
Consider the data :
| Class | 65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 185-205 |
| Frequency | 4 | 5 | 13 | 20 | 14 | 7 | 4 |
The difference of the upper limit of the median class and the lower limit of the modal class is
A0 (B) 19 (C) 20 (D) 38
(C) 20
Explanation:
| Class | Frequency | Cumulative Frequency |
| 65-85 | 4 | 4 |
| 85-105 | 5 | 9 |
| 105-125 | 13 | 22 |
| 125-145 | 20 | 42 |
| 145-165 | 14 | 56 |
| 165-185 | 7 | 63 |
| 185-205 | 4 | 67 |
Here, N/2 = 67/2 = 33.5 which lies in the interval 125 – 145.
Hence, upper limit of median class is 145.
Here, we see that the highest frequency is 20 which lies in 125 – 145.
Hence, the lower limit of modal class is 125.
∴ Required difference = Upper limit of median class – Lower limit of modal class
= 145 – 125 = 20
Hence, option (C) is correct
The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below
| Class | 13.8-14 | 14-14.2 | 14.2-14.4 | 14.4-14.6 | 14.6-14.8 | 14.8-15 |
| Frequency | 2 | 4 | 5 | 71 | 48 | 20 |
The number of athletes who completed the race in less than 14.6 seconds is :
A11 (B) 71 (C) 82 (D) 130
(C) 82
Explanation:
The number of athletes who completed the race in less than 14.6 second= 2 + 4 + 5 + 71 = 82
Hence, option (C) is correct