1. If x1, x2, x3,….., xn are the observations of a given data. Then the mean of the observations will be:
(a) Sum of observations/Total number of observations
(b) Total number of observations/Sum of observations
(c) Sum of observations +Total number of observations
(d) None of the above
Answer: (a) Sum of observations/Total number of observations
Explanation: The mean or average of observations will be equal to the ratio of sum of observations and total number of observations.
xmean=x1+x2+x3+…..+xn/n
2. If the mean of frequency distribution is 7.5 and ∑fi xi = 120 + 3k, ∑fi = 30, then k is equal to:
(a) 40
(b) 35
(c) 50
(d) 45
Answer: (b) 35
Explanation: As per the given question,
Xmean = ∑fi xi /∑fi
7.5 = (120+3k)/30
225 = 120+3k
3k = 225-120
3k= 105
k=35
3. The mode and mean is given by 7 and 8, respectively. Then the median is:
(a) 1/13
(b) 13/3
(c) 23/3
(d) 33
Answer: (c) 23/3
Explanation: Using Empirical formula,
Mode = 3Median – 2 Mean
3Median = Mode + 2Mean
Median = (Mode + 2Mean)/3
Median = [7 + 2(8)]/3 = (7 + 16)/3 = 23/3
4. The mean of the data: 4, 10, 5, 9, 12 is;
(a) 8
(b) 10
(c) 9
(d) 15
Answer: (a) 8
Explanation: mean = (4 + 10 + 5 + 9 + 12)/5 = 40/5 = 8
5. The median of the data 13, 15, 16, 17, 19, 20 is:
(a) 30/2
(b) 31/2
(c) 33/2
(d) 35/2
Answer: (c) 33/2
Explanation: For the given data, there are two middle terms, 16 and 17.
Hence, median = (16 + 17)/2 = 33/2
6. If the mean of first n natural numbers is 3n/5, then the value of n is:
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (c) 5
Explanation: Sum of natural numbers = n(n + 1)/2
Given, mean = 3n/5
Mean = sum of natural numbers/n
3n/5 = n(n + 1)/2n
3n/5 = (n + 1)/2
6n = 5n + 5
n = 5
7. If AM of a, a+3, a+6, a+9 and a+12 is 10, then a is equal to;
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d) 4
Explanation: Mean of AM = 10
(a + a + 3 + a + 6 + a + 9 + a + 12)/5 = 10
5a + 30 = 50
5a = 20
a = 4
8. The class interval of a given observation is 10 to 15, then the class mark for this interval will be:
(a) 11.5
(b) 12.5
(c) 12
(d) 14
Answer: (b) 12.5
Explanation: Class mark = (Upper limit + Lower limit)/2
= (15 + 10)/2
= 25/2
= 12.5
9. If the sum of frequencies is 24, then the value of x in the observation: x, 5,6,1,2, will be;
(a) 4
(b) 6
(c) 8
(d) 10
Answer: (d) 10
Explanation:
Given,
∑fi = 24
∑fi = x + 5 + 6 + 1 + 2 = 14 + x
24 = 14 + x
x = 24 – 14 = 10
10. The mean of following distribution is:
| xi | 11 | 14 | 17 | 20 |
| fi | 3 | 6 | 8 | 7 |
(a) 15.6
(b) 17
(c) 14.8
(d) 16.4
Answer: (d) 16.4
Explanation:
| xi | fi | fixi |
| 11 | 3 | 33 |
| 14 | 6 | 84 |
| 17 | 8 | 136 |
| 20 | 7 | 140 |
| ∑fi = 24 | ∑fi xi = 393 |
xmean = ∑fi xi/∑fi = 393/24 = 16.4
11. Construction of a cumulative frequency table is useful in determining the
(a) mean
(b) median
(c) mode
(d) all the above three measures
Answer: (b) median
Construction of a cumulative frequency table is useful in determining the median.
12. 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
Answer: (b) median
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 median.
13. While computing mean of grouped data, we assume that the frequencies are
(a) centred at the class marks of the classes
(b) evenly distributed over all the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes
Answer: (a) centred at the class marks of the classes
While computing the mean of grouped data, we assume that the frequencies are centred at the class marks of the classes.
14. Consider the following frequency distribution of the heights of 60 students of a class:
| Height (in cm) | 150 – 155 | 155 – 160 | 160 – 165 | 165 – 170 | 170 – 175 | 175 – 180 |
| Number of students | 15 | 13 | 10 | 8 | 9 | 5 |
The sum of the lower limit of the modal class and upper limit of the median class is
(a) 310
(b) 315
(c) 320
(c) 330
Answer: (b) 315
Explanation:
| Height (in cm) | 150 – 155 | 155 – 160 | 160 – 165 | 165 – 170 | 170 – 175 | 175 – 180 |
| Number of students | 15 | 13 | 10 | 8 | 9 | 5 |
| Cumulative frequency | 15 | 28 | 38 | 46 | 55 | 60 |
N/2 = 60/2 = 30
Cumulative frequency nearer and greater than 30 is 38 which corresponds to the class interval 160 – 165.
Thus, median class = 160 – 165
Upper limit of median class = 165
Highest frequency = 15
So, the modal class = 150 – 155
Lower limit of modal class = 150
Therefore, the sum of the lower limit of the modal class and upper limit of the median class = 150 + 165 = 315
15. Consider the following frequency distribution:
| Class | 0 – 5 | 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
Answer: (b) 17.5
Explanation:
Let us write the continuous classes for the given frequency distribution.
| Class | -0.5 – 5.5 | 5.5 – 11.5 | 11.5 – 17.5 | 17.5 – 23.5 | 23.5 – 29.5 |
| Frequency | 13 | 10 | 15 | 8 | 11 |
| Cumulative frequency | 13 | 23 | 38 | 46 | 57 |
N/2 = 57/2 = 28.5
28.5 lies in between the interval 11.5 – 17.5.
Thus, the median class is 11.5 – 17.5.
Therefore, the upper limit of the median class is 17.5.
16. 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 then 14.6 seconds is
(a) 11
(b) 71
(c) 82
(d) 130
Answer: (c) 82
Explanation:
The number of athletes who completed the race in less than 14.6 seconds = 2 + 4 + 5 + 71 = 82
17. Consider the following distribution:
| Marks obtained | Number of students |
| More than or equal to 0 | 63 |
| More than or equal to 10 | 58 |
| More than or equal to 20 | 55 |
| More than or equal to 30 | 51 |
| More than or equal to 40 | 48 |
| More than or equal to 50 | 42 |
the frequency of the class 30-40 is
(a) 3
(b) 4
(c) 48
(d) 51
Answer: (a) 3
Explanation:
The given data can be written as:
| Class | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
| cf | 63 | 58 | 55 | 51 | 48 | 42 |
| f | 5 | 3 | 4 | 3 | 6 | 42 |
Therefore, the frequency of the class 30-40 is 3.
18. The empirical relationship between the three measures of central tendency is
(a) 3 Median = Mode + 2 Mean
(b) 2 Median = Mode + 2 Mean
(c) 3 Median = Mode + Mean
(d) 3 Median = Mode – 2 Mean
Answer: (a) 3 Median = Mode + 2 Mean
The empirical relationship between the three measures of central tendency is 3 Median = Mode + 2 Mean.
19. The ________ of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
(a) Class mark
(b) Class height
(c) Average frequency
(d) Cumulative frequency
Answer: (d) Cumulative frequency
The cumulative frequency of a class is the frequency obtained by adding the frequencies
of all the classes preceding the given class.
20. The method used to find the mean of a given data is(are):
(a) direct method
(b) assumed mean method
(c) step deviation method
(d) all the above
Answer: (d) all the above
The mean for a given data can be calculated using either of the following methods.
Direct method
Assumed mean method
Step deviation method