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Statistics is a branch of mathematics that deals with the study of data collection, analysis, interpretation, presentation, and organization. The field is widely used to acquire a better understanding of data and to focus on its applications. In other words, statistics is the process of collecting, evaluating, and summarising information or data into a mathematical form. Statistics deals with various concepts like data representation, measures of central tendency- mean, median and mode, measures of dispersion, probability etc. there are two statistical approaches in mathematics- descriptive and inferential statistics.
Also read: Statistics Revision Notes
Multiple Choice Questions
Ques. If the mean of frequency distribution is 6.5 and ∑fi xi = 120 + 2k, ∑fi = 20, then k is equal to:
(a)10
(b)15
(c)5
(d)25
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Ans: (c) 5
Explanation: As per the given question,
Xmean = ∑fi xi /∑fi
6.5 = (120+2k)/20
130 = 120+2k
2k = 130-120
2k= 10
k=5
Ques. If x1, x2, x3,….., xn are the observations of a given data. Then the mean of the observations will be:
(a) Total number of observations/Sum of observations
(b) Sum of observations/Total number of observations
(c) Sum of observations +Total number of observations
(d) None of the above
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Ans: (b) 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 (n) of observations.
xmean=x1+x2+x3+…..+xn/n
Ques. The mode and mean are given by 5 and 6, respectively. Then the median is:
(a)17/3
(b)13/3
(c)23/3
(d)33
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Ans: (a) 17/3
Explanation: Using Empirical formula,
Mode = 3Median – 2 Mean
3Median = Mode+2 Mean
Median = (Mode+2Mean)/3
Median = (5+2(6))/3 = (5+12)/3 = 17/3
Ques. The mean of the data: 4, 11, 5, 8, 12 is;
(a)16
(b)8
(c)9
(d)15
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Ans: (b) 8
Explanation: For n values in a set of data namely as x1, x2, x3, … xn, the mean of data is given as:
Therefore, Mean = (4+11+5+8+12)/5 = 40/5 = 8
Ques. The median of the data 11,12, 14, 17, 20, 27 is:
(a)30/2
(b)31/2
(c)33/2
(d)35/2
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Ans: (b) 31/2
Explanation: For the given data, there are two middle terms, 14 and 19.
Hence, median = (14+17)/2 = 31 /2
Ques. Find the mean of frequency distribution, if ∑fi xi = 110, and ∑fi = 10, then k is equal to:
(a) 5
(b) 23
(c) 15
(d) 11
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Ans: (d) 11
Explanation: As per the given question,
Xmean = ∑fi xi /∑fi
= (110)/10
= 11
Therefore, the mean of the frequency distribution is 11.
Ques. The mean and median are given by 4 and 12, respectively. Then the mode is:
(a) 28
(b) 32
(c) 23
(d) 45Maths
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Ans: (a) 28
Explanation: Using Empirical formula,
Mode = 3Median – 2 Mean
= (3 x 12)- 2 x 4
= 36- 8
= 28
Ques. If the mean of first n natural numbers is 4n/6, then the value of n is:
(a) 4
(b) 2
(c) 6
(d) 3
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Ans: (d)
Explanation: Sum of natural numbers = n(n + 1)/2
Given, mean = 4n/6
Mean = sum of natural numbers/n
4n/6 = n(n + 1)/2n
4n/6 = (n + 1)/2
8n = 6n + 6
n = 3
Ques. If the mean of a, a+2, a+4, a+8 and a+11 is 10, then a is equal to;
(a) 1
(b) 2
(c) 5
(d) 4
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Ans: (c) 5
Explanation: Mean = 10
(a + a + 2 + a + 4 + a + 8 + a + 11)/5 = 10
5a + 25 = 50
5a = 25
a = 5
Ques. The class interval of a given observation is 5 to 10, then the class mark for this interval will be:
(a) 7.5
(b) 13.5
(c) 8
(d) 14.5
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Ans: (a) 7.5
Explanation: Class mark = (Upper limit + Lower limit)/2
= (10 + 5)/2
= 15/2
= 7.5
Ques. If the sum of frequencies is 30, then the value of x in the observation: x, 5,6,1,2, will be;
(a) 4
(b) 16
(c) 9
(d) 10
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Ans: (b) 16
Explanation:
Given,
∑fi = 30
∑fi = x + 5 + 6 + 1 + 2 = 14 + x
30 = 14 + x
x = 30 – 14 = 16
Ques. Cumulative frequency curve is also called
(a) ogive
(b) histogram
(c) bar graph
(d) median
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Ans: (a) ogive
Explanation: Ogive
A cumulative frequency curve is a graphical representation of the cumulative frequencies of a given data. It is called ogive. An ogive can either be less than ogive or more than ogive.
Less Than Ogive
Less than ogive is a graphical representation of the frequency table, where the cumulative frequency starts from the total of the frequency and moves on to the lowest frequency.
More than Ogive
More than ogive is a graphical representation of the frequency table, where the cumulative frequency starts from the lowest frequency and moves on to the total of the frequencies.
Ques. If the sum of frequencies is 36, then the value of x in the observation: x, 4,6,1,5, will be;
(a)4
(b)10
(c)9
(d) 6
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Ans: (b) 10
Explanation:
Given,
∑fi = 36
∑fi = x+4+6+1+5=16+x
36 = 16+x
x=36-16 = 10
Ques. The relationship between the measure of central tendency- mean, median and mode for a moderately skewed distribution is given as:
(a) mode = 3 median – 2 mean
(b) mode = median – 2 mean
(c) mode = 2 median – 3 mean
(d) mode = median – mean
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Ans: (a)mode = 3 median – 2 mean
Explanation: the relation between or empirical formula for the measure of central tendency- mean, median and mode is given as:
Mode = 3 median – 2 mean
Mean= (3 median-mode)/2
Median= (mode- 2 mean)/3
Ques. Mode and mean of the data are 6k and 12k. Median of the data is
(a) 12k
(b) 10k
(c) 15k
(d) 16k
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Ans: (b) 10k
Explanation:
The relation between mean, median and mode is given as:
Mode = 3 median – 2 mean
Therefore, 6k = 3 median – 2 × 12k
3 median = 24k + 6k
Median= 30k/3
⇒ Median = 10k.
Ques. Mean of n numbers x1, x2, x3… xn is m. If xn is replaced by x, then new mean is
(a) m – xn + x
(b) (m−xn+xn)/n
(c) (n−1)m+xn
(d) (nm−xn+xn)/n
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Ans: (d) (nm−xn+xn)/n
Explanation:
Ques. The mean of following distribution is
| xi | 11 | 14 | 17 | 20 |
| fi | 2 | 6 | 8 | 5 |
(a)12.6
(b)15
(c)16.2
(d)18.4
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Ans: (c) 16.2
Explanation:
| xi | fi | fixi |
| 11 | 2 | 22 |
| 14 | 6 | 84 |
| 17 | 8 | 136 |
| 20 | 5 | 100 |
| ∑fi = 21 | ∑fi xi=342 |
xmean = ∑fi xi/∑fi
= 342/21 = 16.2
Ques. 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) 315
(b) 310
(c) 320
(c) 330
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Ans: (a) 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 = (15+ 13+ 10+ 8+ 9+ 5)/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
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