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Maths M.C.Q (1 Marks)

Mean, Median, Mode of Grouped Data · CBSE STD 10 (CBSE - English Medium). 34 questions.

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M.C.Q (1 Marks)

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34 questions · timed · auto-graded

MCQ 11 Mark
Median =?
  • ✓
    $\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
  • B
    $\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\text{cf}-\frac{\text{N}}{2}\Big)}{\text{f}}\end{Bmatrix}$
  • C
    $\text{l}-\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
  • D
    None of these.
Answer
Correct option: A.
$\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$\text{l+}\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
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MCQ 21 Mark
The relation between mean, mode and median is:
  • A
    Mode = (3 × mean) − (2 × median)
  • ✓
    Mode = (3 × median) − (2 × mean)
  • C
    Mode = (3 × mean) − (2 × mode)
  • D
    Mode = (3 × median) − (2 × mode)
Answer
Correct option: B.
Mode = (3 × median) − (2 × mean)
Mode = (3 × median) − (2 × mean)
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MCQ 31 Mark
Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer: use the following code:
Assertion (A)
Consider the following frequency distribution:
Class interval
3-6
6-9
9-12
12-15
15-18
18-21
Frequency
2
5
21
23
10
12
The mode of the above data is 12.4.
Reason (R)
The value of the variable which occurs most often is the mode.
  • A
    Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
  • ✓
    Both Assertion (A) and Reason (R) are true but Reason (R) is a not a correct explanation of Assertion (A).
  • C
    Assertion (A) is true and Reason (R) is false.
  • D
    Assertion (A) is false and Reason (R) is true
Answer
Correct option: B.
Both Assertion (A) and Reason (R) are true but Reason (R) is a not a correct explanation of Assertion (A).
Reason (R) is true.
Maxximum frequency = 23
Hence, modakl class is 12-15
Now, mode $=\text{x}_\text{k}+\text{h}\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
$=12+3\Big\{\frac{(23-21)}{(2(23)-21-10)}\Big\}$
$=12+3\times\frac{2}{15}$
$=12+0.4$
$=12.4$
Thus, Assertion (A) and Reason (R) are both true but Reason (R) rs nor !he eereee explananoon of Assertion (A).
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MCQ 41 Mark
Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer: use the following code:
Assertion (A) Reason (R)
If the median and mode of a frequency distribution are 150 and 154 respectively, then its mean is 148. Mean, median and mode of a frequency distribution are related as: mode = 3median - 2mean
  • ✓
    Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
  • B
    Both Assertion (A) and Reason (R) are true but Reason (R) is a not a correct explanation of Assertion (A).
  • C
    Assertion (A) is true and Reason (R) is false.
  • D
    Assertion (A) is false and Reason (R) is true.
Answer
Correct option: A.
Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Reason (R) is true.
Using the relation given in (R), we have
Median = 150
Mode = 154
Mode = 3Median - 2Meen
Hence, mean $=\frac{3\text{Median}-\text{Mode}}{2}=\frac{3(150)-154}{2}=\frac{450-154}{2}=\frac{296}{2}=148$
Thus, Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
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MCQ 51 Mark
Which of the following measures of central tendency is influenced by extreme values?
  • ✓
    Mean
  • B
    Median
  • C
    Mode
  • D
    None of these
Answer
Correct option: A.
Mean
Since mean is the average of all observations, it is influenced by extreme values.
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MCQ 61 Mark
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
    16.5
  • B
    18.5
  • C
    18
  • ✓
    17.5
Answer
Correct option: D.
17.5
The given series is in indusive from. comverting in to exclusive from and preparing cumulative frequency tabla, we get
Class Frequency Cumulative frequency
-0.5-5.5 13 13
5.5-11.5 10 23
11.5-17.5 15 38
17.5-23.5 8 46
23.5-29.5 11 57
Here, $\text{N}=57\Rightarrow\frac{\text{N}}{2}=28.5$

The cumulative frequency just greater than 28.5 is 38.

Hence, the median class is 11.5-17.5

$\therefore$ Upper limit of the median class = 17.5
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MCQ 71 Mark
If the 'less than type' ogive and 'more than type' ogive intersect each other at (20.5, 15.5) then the median of the given data is:
  • A
    5.5
  • B
    15.5
  • ✓
    20.5
  • D
    36.0
Answer
Correct option: C.
20.5
Since the abscissa of the point of intersection of both ihe ogives gives the median, we have median: 20.5
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MCQ 81 Mark
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
  • ✓
    Median
  • C
    Mode
  • D
    None of these
Answer
Correct option: B.
Median
Median is given by the abscissa of the point of intersection of the Less than Type and More than Type cumulative frequency curves.
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MCQ 91 Mark
Which of the following is not a measure of central tendency?
  • A
    Mean
  • B
    Mode
  • C
    Median
  • ✓
    Standard deviation.
Answer
Correct option: D.
Standard deviation.
Range is not a measure of central tendency.
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MCQ 101 Mark
Mode =?
  • A
    $\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_{\text{k}-1}-\text{f}_\text{k})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
  • ✓
    $\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
  • C
    $\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(\text{f}_\text{k}-2\text{f}_{\text{k}-1}-\text{f}_{\text{k}-1})}\Big\}$
  • D
    $\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(\text{f}_{\text{k}}-\text{f}_{\text{k}-1}-2\text{f}_{\text{k}+1})}\Big\}$
Answer
Correct option: B.
$\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
$\text{x}_\text{k}+\text{h}.\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
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MCQ 111 Mark
The median of first 8 prime numbers is:
  • A
    7
  • B
    9
  • C
    11
  • ✓
    13
Answer
Correct option: D.
13
First 8 prime numbers are are follows:
2, 3, 5, 7, 11, 13, 17, 19
N = 8 (even)
$\therefore$ Median $=\frac{\Big(\frac{8}{2}\Big)^{\text{th}}\text{value}+\Big(\frac{8}{2}+1\Big)^{\text{th}}\text{value}}{2}$
$=\frac{4^{\text{th}}\text{value }+\ 5^{\text{th}}\text{Value}}{2}$
$=\frac{7+11}{2}$
$=\frac{18}{2}$
$=9$
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MCQ 121 Mark
The mean of 2, 7, 6 and x is 15 and the mean of 18, 1, 6, x and y is 10. What is the value of y?
  • A
    5
  • B
    10
  • ✓
    20
  • D
    30
Answer
Correct option: C.
20
Mean of 2, 7, 6 and x = 5$\Rightarrow\frac{2+7+6+\text{x}}{4}=5$
$\Rightarrow15+\text{x}=20$
$\Rightarrow\text{x}=5$
Mean of 18, 1, 6, x and y = 10
$\Rightarrow\frac{18+1+6+\text{x}+\text{y}}{5}=10$
$\Rightarrow\frac{18+1+6+\text{5}+\text{y}}{5}=10$
$\Rightarrow30+\text{y}=50$
$\Rightarrow\text{y}=20$
Note: Question modified
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MCQ 131 Mark
If the mean of a data is 27 and its median is 33. Then, the mode is:
  • A
    30
  • B
    43
  • ✓
    45
  • D
    47
Answer
Correct option: C.
45
Mean = 27Median = 33
Mode = 3median - 2Mean
= 3 × 33 - 2 × 27
= 99 - 54
= 45
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MCQ 141 Mark
For finding the mean by using the formula, $\bar{\text{x}}=\text{A}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big),$ we have $u_i =?$
  • A
    $\frac{(\text{A}-\text{x}_\text{i})}{\text{h}}$
  • ✓
    $\frac{\text{(x}_{\text{i}}-\text{A})}{\text{h}}$
  • C
    $\frac{(\text{A}+\text{x}_\text{i})}{\text{h}}$
  • D
    $\text{h}(\text{x}_\text{i}-\text{A})$
Answer
Correct option: B.
$\frac{\text{(x}_{\text{i}}-\text{A})}{\text{h}}$
By formula method,
$\bar{\text{x}}=​​​​\text{A}+\text{h}\Big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\Big),$
where $\text{u}_\text{i}=\frac{(\text{x}_\text{i}-\text{A})}{\text{h}}$
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MCQ 151 Mark
The mean and mode of a frequency distribution are 28 and 16 respectively. The median is:
  • A
    22
  • B
    23.5
  • ✓
    24
  • D
    24.5
Answer
Correct option: C.
24
mean = 28Mode = 16
Mode = 3median - 2mean
Hence, median $=\frac{\text{Mode + 2Mean}}{3}$
$=\frac{16+2(28)}{3}$
$=\frac{16+56}{3}$
$=\frac{72}{3}$
$=24$
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MCQ 161 Mark
The cumulative frequency table is useful in determining the:
  • A
    Mean
  • ✓
    Median
  • C
    Mode
  • D
    All of these
Answer
Correct option: B.
Median
The cumulative frequency table is useful in determining the medan.
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MCQ 171 Mark
If the mean and median of a set of number are 8.9 and 9 respectively, then the mode will be:
  • A
    7.2
  • B
    8.2
  • ✓
    9.2
  • D
    10.2
Answer
Correct option: C.
9.2
Mean = 8.9
Median = 9
Mode = 3median - 2mean
= 3 × 9 - 2 × 8.9
= 27 - 27.8
= 9.2
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MCQ 181 Mark
The median of a frequency distribution is found graphically with the help of:
  • A
    A histogram.
  • B
    A frequency curve.
  • C
    A frequency polygon.
  • ✓
    Ogives
Answer
Correct option: D.
Ogives
ogives are used to determine the median of a frequency distribution.
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MCQ 191 Mark
While computing the mean of the grouped data, we assume that the frequencies are:
  • A
    Evenly distributed over the classes.
  • ✓
    Centred at the class marks of the classes.
  • C
    Centred at the lower limits of the classes.
  • D
    Centred at the upper limits of the classes.
Answer
Correct option: B.
Centred at the class marks of the classes.
While computing the mean of the grouped data. we assume that the frequencies are centred at the class marks of the classes.
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MCQ 201 Mark
Consider the following distribution:
Class 0-5 5-10 10-15 15-20 20-25
Frequency 10 15 12 20 9
The sum of the lower limits of the median class and the modal class is:
  • A
    15
  • ✓
    25
  • C
    30
  • D
    35
Answer
Correct option: B.
25
Class interval Frequency frequency
0-10 10 10
5-10 15 25
10-15 15 37
15-20 20 57
20-25 9 66
Here, $\text{N}=66\Rightarrow\frac{\text{N}}{2}=33$

The cumulative frequency just greater than 33 is 37.

Hence, the median class is 10-15

$\therefore$ Lower limit of median class = 10

Class having maximum frequency is the modal class.

Here, maximum frequency = 20

$\therefore$ Lower limit of modal class = 15

$\therefore$ Required sum = 10 + 15 = 25
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MCQ 211 Mark
Consider the following frequency distribution:
Class 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 3 9 15 30 18 5
The modal class is:
  • A
    10-20
  • B
    20-30
  • ✓
    30-40
  • D
    50-60
Answer
Correct option: C.
30-40
Class having maxiroom frequency is the modal class.
Here, maximum frequency = 30
Hence. the modal class is 30-40.
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MCQ 221 Mark
If $x_i$ s are the midpoints of the class intervals of a grouped data, $f_i$ s are the corresponding frequency and $\bar{\text{x}}$ is the mean, then $\sum\text{f}_{\text{i}}(\text{x}_\text{i}-\bar{\text{x}})=?$
  • A
    $1$
  • ✓
    $0$
  • C
    $-1$
  • D
    $2$
Answer
Correct option: B.
$0$
For a grouped data
$\sum\text{f}_\text{i}(\text{x}_\text{i}-\bar{\text{x}})=0$
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MCQ 231 Mark
Which one of the following measures is determined only after the construction of cumulative frequency distribution?
  • A
    Mean
  • ✓
    Median
  • C
    Mode
  • D
    None of these
Answer
Correct option: B.
Median
The cumulative frequency table is useful in determining the median.
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MCQ 241 Mark
The mode of a frequency distribution is obtained graphically from:
  • A
    A frequency curve.
  • B
    A frequency polygon.
  • ✓
    A histogram.
  • D
    An ogive.
Answer
Correct option: C.
A histogram.
Mode can be obtained graphically from a histogram.
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MCQ 251 Mark
The median and mode of a frequency distribution are 26 and 29 respectively. Then, the mean is:
  • A
    27.5
  • ✓
    24.5
  • C
    28.4
  • D
    25.8
Answer
Correct option: B.
24.5
Median = 26
Mode = 29
Mode = 3median - 2mean
Hence, Mean $=\frac{3\text{Median}-\text{Mode}}{2}$
$=\frac{3(26)-29}{2}$
$=\frac{78-29}{2}$
$=\frac{49}{2}$
$=24.5$
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MCQ 261 Mark
Look at the cumulative frequency distribution table given below:
Monthly income
Number of families
More that Rs 10000
100
More that Rs 14000
85
More that Rs 18000
69
More that Rs 20000
50
More that Rs 25000
37
More that Rs 30000
15
Number of families having income range Rs 20000 to Rs 25000 is:
  • A
    19
  • B
    16
  • ✓
    13
  • D
    22
Answer
Correct option: C.
13
Number of families having income more than Rs. 20000 = 50

Number of families having income more than Rs. 25000 = 37

Hence, number of families having Income range 20000 to 25000 = 50 - 37 = 13
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MCQ 271 Mark
For a symmetrical frequency distribution, we have:
  • A
    Mean < mode < median
  • B
    Mean > mode > median
  • ✓
    Mean = mode = median
  • D
    Mode $=\frac{1}{2}$ (mean + median)
Answer
Correct option: C.
Mean = mode = median
For a symmetrical distribution, we haveMean = mode = median
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MCQ 281 Mark
In the formula, $\bar{\text{x}}=\Big\{\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big\}$ for finding the mean of the grouped data, the $d_i\ 's$ are the deviations from $A$ of:
  • A
    Lower limits of the classes
  • B
    Upper limits of the classes
  • ✓
    Midpoints of the classes
  • D
    None of these
Answer
Correct option: C.
Midpoints of the classes
$d_1$ 's are the deviations from $A$ of midpoints of the classes
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MCQ 291 Mark
Consider the frequency distribution of the heights of 60 students of a class:
Height (in cm) No. of Students Cumulative Frequency
150-155 16 16
155-160 12 28
160-165 9 37
165-170 7 44
170-175 10 54
175-180 6 60
The sum of the lower limit of the modal class and the upper limit of the median class is:
  • A
    310
  • ✓
    315
  • C
    320
  • D
    330
Answer
Correct option: B.
315
Class having maximum frequency is the modal class.
Hence, modal class: 150- 155.
$\therefore$ Lower Iimit of the modal class = 150
$\text{N}=60\Rightarrow\frac{\text{N}}{2}=30$
The cumulative frequency just greater than 30 is 37.
Hence, the median class is 160-165.
$\therefore$ Upper limit of the median class = 165
Required sum= 150 + 165 = 315
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MCQ 301 Mark
The mean of 20 numbers is zero. of them, at the most, how many may be greater than zero?
  • A
    0
  • B
    1
  • C
    10
  • ✓
    19
Answer
Correct option: D.
19
Mean of 20 numbers = 0Hence, sum of 20 numbers = 0 × 20 = 0
Now, the mean can be zero if
Sum of 10 numbers is (S) and the sum of remaining 10 numbers is (-S),
Sum of 11 numbers is (S) and the sum of remaining 9 numbers is (-S),
Sum of 19 nun'bers is (S) and the 20th number is (-S), then their sum is zero.
So, at the most 19 numbers can be greater than zero.
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MCQ 311 Mark
Which of the following cannot be determined graphically?
  • ✓
    Mean
  • B
    Median
  • C
    Mode
  • D
    None of these
Answer
Correct option: A.
Mean
Mean cannot be determined graphically.
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MCQ 321 Mark
Consider the following table:
Class interval
10-14
14-18
18-22
22-26
26-30
Frequency
5
11
16
25
19
The mode of the above data is:
  • A
    23.5
  • B
    24
  • ✓
    24.4
  • D
    25
Answer
Correct option: C.
24.4
Maximum frequency = 25
Hence, modal class is 22-26
Now, mode $=\text{x}_\text{k}+\text{h}\Big\{\frac{(\text{f}_\text{k}-\text{f}_{\text{k}-1})}{(2\text{f}_\text{k}-\text{f}_{\text{k}-1}-\text{f}_{\text{k}+1})}\Big\}$
$22+4\Big\{\frac{(25-16)}{(2(25)-16-19)}\Big\}$
$=22+4\times\frac{9}{15}$
$=22+2.4$
$=24.4$
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MCQ 331 Mark
If the median of the data 4, 7, x - 1, x - 3, 16, 25, written in ascending order, is 13 then x is equal to
  • A
    13
  • B
    14
  • ✓
    15
  • D
    16
Answer
Correct option: C.
15
Data in ascending odhar:
4, 7, x - 1, x - 3, 16, 25
N = 6 (even)
$\therefore$ median $=\frac{\Big(\frac{6}{2}\Big)^{\text{th}}\text{value}+\Big(\frac{6}{2}+1\Big)^{\text{th}}\text{value}}{2}$
$\therefore13=\frac{3^{\text{rd}}\text{value}+4^{\text{th}}\text{value}}{2}$
$\therefore26-2\text{x}-4$
$\therefore2\text{x}=30$
$\therefore\text{x}=15$
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MCQ 341 Mark
Look at the frequency distribution table given below:
Class interval
35-45
45-55
55-65
65-75
Frequency
8
12
20
10
The median of the above distribution is:
  • A
    56.5
  • B
    57.5
  • ✓
    58.5
  • D
    59
Answer
Correct option: C.
58.5
 
Class intervalClass interval
FrequencyFrequency
Cumulative frequencyCumulative frequency
35-45
8
8
45-55
12
20
55-65
20
40
65-75
10
50
Here, $\text{N}=50\Rightarrow\frac{\text{N}}{2}=25$
The cumulaove frequency just greater than 25 is 40.
Hence, median class is 55-65.
Now, median $=\text{l}+\begin{Bmatrix}\text{h}\times\frac{\Big(\frac{\text{N}}{2}-\text{cf}\Big)}{\text{f}}\end{Bmatrix}$
$=55+\Big\{10\times\frac{(25-20)}{20}\Big\}$
$=55+2.5$
$=57.5$
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