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MATHS M.C.Q (1 Marks)

Statistics · UP Board (UPMSP) STD 10 (UPMSP - English Medium). 110 questions.

Questions · Page 2 of 3

M.C.Q (1 Marks)

MCQ 511 Mark
The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data gives its
  • A
    mean
  • ✓
    median
  • C
    mode
  • D
    all the three above
Answer
Correct option: B.
median
b
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MCQ 521 Mark
Consider the following frequency distribution:
Class:0-56-1112-1718-2324-29
Frequency:131015811
The upper limit of the median class is
  • A
    17
  • ✓
    17.5
  • C
    18
  • D
    18.5
Answer
Correct option: B.
17.5
b
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MCQ 531 Mark
For the following distribution:
Below:102030405060
Number of students:31227577580
the modal class is
  • A
    10 - 20
  • B
    20 - 30
  • ✓
    30 - 40
  • D
    50 - 60
Answer
Correct option: C.
30 - 40
c
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MCQ 541 Mark
For the following distribution:
Class:0-55-1010-1515-2020-25
Frequency:101512209
the sum of the lower limits of the median and modal class is
  • A
    15
  • ✓
    25
  • C
    30
  • D
    35
Answer
Correct option: B.
25
b
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MCQ 551 Mark
In the formula $\bar{X}=a+h\left(\frac{1}{N} \sum f_i u_i\right)$,for finding the mean of grouped frequency distribution $u_i=$
  • A
    $\frac{x_i+a}{h}$
  • B
    $h\left(x_i-a\right)$
  • ✓
    $\frac{x_i-a}{h}$
  • D
    $\frac{a-x_i}{h}$
Answer
Correct option: C.
$\frac{x_i-a}{h}$
c
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MCQ 561 Mark
While computing mean of grouped data, we assume that the frequencies are
  • A
    evenly distributed over all the classes
  • ✓
    centred at the class marks of the classes
  • C
    centred at the upper limit of the classes
  • D
    centred at the lower limit of the classes.
Answer
Correct option: B.
centred at the class marks of the classes
b
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MCQ 571 Mark
If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by
  • A
    2
  • B
    1.5
  • C
    1
  • ✓
    0.5
Answer
Correct option: D.
0.5
d
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MCQ 581 Mark
If $u_i=\frac{x_i-25}{10}, \Sigma f_i u_i=20, \Sigma f_i=100$, then $\bar{x}=$
  • A
    23
  • B
    24
  • ✓
    27
  • D
    25
Answer
Correct option: C.
27
c
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MCQ 591 Mark
Mean of a certain number of observations is $\bar{x}$.If each observation is divided by $m(m \neq 0)$ and increased by in, then the mean of new observation is
  • ✓
    $\frac{\bar{x}}{m}+n$
  • B
    $\frac{\bar{x}}{n}+m$
  • C
    $\bar{x}+\frac{n}{m}$
  • D
    $\bar{x}+\frac{m}{n}$
Answer
Correct option: A.
$\frac{\bar{x}}{m}+n$
a
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MCQ 601 Mark
If the mean of observations $x_1, x_2, \ldots, x_n$ is $\bar{x}$, then the mean of $x_1+a, x_2+a, \ldots, x_n+a$ is
  • A
    $a \bar{x}$
  • B
    $\bar{x}-a$
  • ✓
    $\bar{x}+a$
  • D
    $\frac{\bar{x}}{a}$
Answer
Correct option: C.
$\bar{x}+a$
c
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MCQ 621 Mark
If mode of a series exceeds its mean by 12, then mode exceeds the median by
  • A
    4
  • ✓
    8
  • C
    6
  • D
    10
Answer
Correct option: B.
8
b
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MCQ 641 Mark
If the difference of mode and median of a data is 24, then the difference of median and mean is
  • ✓
    12
  • B
    24
  • C
    8
  • D
    36
Answer
Correct option: A.
12
a
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MCQ 651 Mark
The mean of first n odd natural numbers is $\frac{n^2}{81}$, then n =
  • A
    9
  • ✓
    81
  • C
    27
  • D
    18
Answer
Correct option: B.
81
b
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MCQ 661 Mark
The mean of first n odd natural number is
  • A
    $\frac{n+1}{2}$
  • B
    $\frac{n}{2}$
  • ✓
    n
  • D
    $n^2$
Answer
Correct option: C.
n
c
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MCQ 671 Mark
The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is
  • A
    25
  • B
    18
  • ✓
    20
  • D
    22
Answer
Correct option: C.
20
c
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MCQ 691 Mark
The mean of n observations is $\bar{x}$ . If the first observation is increased by 1, the second by 2, the third by 3, and so on, then the new mean is
  • A
    $\bar{x}+(2 n+1)$
  • ✓
    $\bar{x}+\frac{n+1}{2}$
  • C
    $\bar{x}+(n+1)$
  • D
    $\bar{x}-\frac{n+1}{2}$
Answer
Correct option: B.
$\bar{x}+\frac{n+1}{2}$
b
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MCQ 701 Mark
If the mean of 6, 7, x, 8, y, 14 is 9, then
  • A
    x + y = 21
  • ✓
    x + y = 19
  • C
    x - y = 19
  • D
    x - y = 21
Answer
Correct option: B.
x + y = 19
b
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MCQ 711 Mark
If the mean of a frequency distribution is 8.1 and $\Sigma f_i x_i=132+5 k, \Sigma f_i=20$, then $k=$
  • A
    3
  • B
    4
  • C
    5
  • ✓
    6
Answer
Correct option: D.
6
d
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MCQ 721 Mark
The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m - 1 and median q. Then, p + q =
  • A
    4
  • B
    5
  • C
    6
  • ✓
    7
Answer
Correct option: D.
7
d
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MCQ 731 Mark
If the mode of the data: 16, 15, 17, 16, 15, x, 19, 17, 14 is 15, then x =
  • ✓
    15
  • B
    16
  • C
    17
  • D
    19
Answer
Correct option: A.
15
a
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MCQ 741 Mark
If the mode of the data: 64, 60, 48, x, 43, 48, 43, 34 is 43, then x + 3 =
  • A
    44
  • B
    45
  • ✓
    46
  • D
    48
Answer
Correct option: C.
46
c
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MCQ 761 Mark
If the median of the data: 6, 7, x - 2 x, 17, 20, written in ascending order, is 16 Then x =
  • A
    15
  • B
    16
  • ✓
    17
  • D
    18
Answer
Correct option: C.
17
c
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MCQ 771 Mark
If the median of the data: 24, 25, 26, x + 2 x + 3 30, 31, 34 is 27.5, then x =
  • A
    27
  • ✓
    25
  • C
    28
  • D
    30
Answer
Correct option: B.
25
b
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MCQ 781 Mark
If the arithmetic mean of x, x + 3 x + 6 x + 9 and x + 12 is 10, the x =
  • A
    1
  • B
    2
  • C
    6
  • ✓
    4
Answer
Correct option: D.
4
d
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MCQ 791 Mark
The mean of a discrete frequency distribution $x_i / f_i ; i=1,2, \ldots, n$ is given by
  • ✓
    $\frac{\sum f_i x_i}{\sum f_i}$
  • B
    $\frac{1}{n} \sum_{i=1}^n f_i x_i$
  • C
    $\frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n x_i}$
  • D
    $\frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n i}$
Answer
Correct option: A.
$\frac{\sum f_i x_i}{\sum f_i}$
a
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MCQ 801 Mark
The relationship between mean, median and mode for a moderately skewed distribution is
  • A
    Mode = 2 Median - 3 Mean
  • B
    Mode = Median - 2 Mean
  • C
    Mode = 2 Median - Mean
  • ✓
    Mode = 3 Median - 2 mean
Answer
Correct option: D.
Mode = 3 Median - 2 mean
d
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MCQ 811 Mark
If the mean of the following distribution is 2.6, then the value of y is
Variable (x) :12345
Frequency :45y12
  • A
    3
  • ✓
    8
  • C
    13
  • D
    24
Answer
Correct option: B.
8
b
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MCQ 821 Mark
One of the methods of determining mode is
  • A
    Mode = 2 Median - 3 Mean
  • B
    Mode = 2 Median + 3 Mean
  • ✓
    Mode = 3 Median - 2 Mean
  • D
    Mode = 3 Median + 2 Mean
Answer
Correct option: C.
Mode = 3 Median - 2 Mean
c
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MCQ 831 Mark
The mean of in observations is $\bar{X}$. If the first item is increased by 1, second by 2 and so on, then the new mean is
  • A
    $\bar{X}+n$
  • B
    $\bar{x}+\frac{n}{2}$
  • ✓
    $\bar{X}+\frac{n+1}{2}$
  • D
    none of these
Answer
Correct option: C.
$\bar{X}+\frac{n+1}{2}$
c
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MCQ 841 Mark
Mode is
  • A
    least frequent value
  • B
    middle most value
  • ✓
    most frequent value
  • D
    none of these
Answer
Correct option: C.
most frequent value
c
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MCQ 851 Mark
The mode of a frequency distribution can be determined graphically from
  • ✓
    Histogram
  • B
    Frequency polygon
  • C
    Ogive
  • D
    Frequency curve
Answer
Correct option: A.
Histogram
a
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MCQ 861 Mark
The median of a given frequency distribution is found graphically with the help of
  • A
    Histogram
  • B
    Frequency curve
  • C
    Frequency polygon
  • ✓
    Ogive
Answer
Correct option: D.
Ogive
d
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MCQ 871 Mark
Which of the following cannot be determined graphically?
  • ✓
    Mean
  • B
    Median
  • C
    Mode
  • D
    none of these
Answer
Correct option: A.
Mean
a
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MCQ 881 Mark
For a frequency distribution, mean, median and mode are connected by the relation
  • A
    Mode = 3 Mean - 2 Median
  • B
    Mode = 2 Median - 3 Mean
  • ✓
    Mode = 3 Median - 2 Mean
  • D
    Mode = 3 Median + 2 Mean
Answer
Correct option: C.
Mode = 3 Median - 2 Mean
c
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MCQ 891 Mark
The arithmetic mean of 1, 2, 3, ..., n is
  • ✓
    $\frac{n+1}{2}$
  • B
    $\frac{n-1}{2}$
  • C
    $\frac{n}{2}$
  • D
    $\frac{n}{2}+1$
Answer
Correct option: A.
$\frac{n+1}{2}$
a
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MCQ 901 Mark
The algebraic sum of the deviations of a frequency distribution from its mean is
  • A
    always positive
  • B
    always negative
  • ✓
    $0$
  • D
    a non-zero number
Answer
Correct option: C.
$0$
c
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MCQ 911 Mark
Which of the following is not a measure of central tendency
  • A
    Mean
  • B
    Median
  • C
    Mode
  • ✓
    Standard deviation
Answer
Correct option: D.
Standard deviation
d
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MCQ 921 Mark
The mean of the data $x+a, x+2 a,(x+3 a), \ldots, x+(2 n+1) a$ is
  • ✓
    $x+(n+1) a$
  • B
    $x+(n-1) a$
  • C
    $x+(n+2) a$
  • D
    $x+n a$
Answer
Correct option: A.
$x+(n+1) a$
(a)
Let $\bar{X}$ be the required mean. Then,
$\bar{X}=\frac{(x+a)+(x+2 a)+(x+3 a)+\ldots+(x+(2 n+1) a)}{2 n+1}$
$\Rightarrow \quad \bar{X}=\frac{x(2 n+1)+a\{1+2+3+\ldots+(2 n+1)\}}{2 n+1}$
$\Rightarrow \quad \bar{X}=x+\frac{a}{2 n+1}\left\{\frac{2 n+1}{2}(1+2 n+1)\right\}=x+a(n+1)$
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MCQ 931 Mark
While finding the mean of the grouped data by using the formula $\bar{X}=a+\frac{1}{N} \sum f_1 d_i$ $d_i$ 's are the diviations from a of
  • ✓
    the mid-values of the classes
  • B
    lower limits of the classes
  • C
    upper limits of the classes
  • D
    frequencies of the class marks
Answer
Correct option: A.
the mid-values of the classes
(a)
In the given formula, $d_i=x_i-a$, where $x_i$ are the mid-values of the classes
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MCQ 941 Mark
If the mean and median of a data are 12 and 15 respectively, then its mode is
  • A
    13.5
  • ✓
    21
  • C
    6
  • D
    14
Answer
Correct option: B.
21
(b)
We have, Mean = 12 and Median = 15
∴ Mode = 3 Median-2 Mean ⇒ Mode = 3 * 15 - 2 * 12 = 21 .
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MCQ 951 Mark
If the value of cach observation of a statistical data consisting of n observations is increased by 3, then the mean of the data
  • A
    remains unchanged
  • ✓
    increases by 3
  • C
    increases by 6
  • D
    increases by 3n
Answer
Correct option: B.
increases by 3
(b)
Let $x_1, x_2, \ldots, x_n$ be $n$ values of a variable X and Y be a new variable taking values $y_1, y_2, \ldots, y_n$ such that $y_i=a x_1+b ; i=1,2, \ldots, n$. Then $\bar{Y}=a \bar{X}+b$.
Here, a = 1 and b = 3. Therefore, $\bar{Y}=\bar{X}+3$. Hence, the mean is increased by 3.
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MCQ 961 Mark
Which computing the mean of grouped data, it is assumed that the frequencies are
  • A
    centred at the lower limits of the classes
  • B
    centred at the upper limits of the classes
  • ✓
    centred at the class marks of the classes
  • D
    evenly distributed over all the classes.
Answer
Correct option: C.
centred at the class marks of the classes
(c)
The mean $\bar{X}$ of the grouped data is given by
$\bar{X}=\frac{1}{N} \sum f_1 x_1$ where $x_i^{\prime} s$ are the mid-values of the class-intervals.
Therefore, class frequencies are centred at the class marks of the classes.
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MCQ 971 Mark
The mean and median of the data a, b and c are 50 and 35 respectively, where a < b < c If c - a = 55 then b - a =
  • A
    8
  • B
    7
  • C
    3
  • ✓
    5
Answer
Correct option: D.
5
(d)
It is given that a < b < c Therefore, median = b. But, it is given that the median is 35.
Therefore, b = 35.
Mean of a, b and c is 50.
$\therefore \quad \frac{a+b+c}{3}=50 \Rightarrow a+b+c=150 \Rightarrow a+35+c=150 \Rightarrow a+c=115$
Thus, we have c + a = 115 and c - a =55 Rightarrow a=30 and c = 85
Hence, b - a = 35 - 30 = 5
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MCQ 981 Mark
3 If the arithmetic mean of 2, 4, 6, 8, 3 and 7 is 5, then the arithmetic mean of 1002, 1004, 1006, 1008, 1003 and 1007 is
  • ✓
    1005
  • B
    1004
  • C
    1008
  • D
    1008
Answer
Correct option: A.
1005
(a)
It is given that AM of 2, 4, 6, 8, 3, 7 is 5. Therefore, by increasing each observation by 1000, the AM also increases by 1000. Hence, the AM of 1002, 1004, 1006, 1008, 1003 and 1007 is 1005.
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MCQ 991 Mark
If the arithmetic mean of first n natural numbers is 15, then n is equal to
  • A
    14
  • B
    15
  • ✓
    29
  • D
    30
Answer
Correct option: C.
29
(c)
It is given that
$15=\frac{1+2+3+\ldots+n}{n} \Rightarrow 15=\frac{n(n+1)}{2 n} \Rightarrow 15=\frac{n+1}{2} \Rightarrow n+1=30 \Rightarrow n=29$
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MCQ 1001 Mark
For some data $x_1, x_2, \ldots, x_n$ with respective frequencies $f_1, f_2, \ldots, f_n$, the value of $\sum_{i=1}^n f_i\left(x_1-\bar{X}\right)$ is equal to
  • A
    $n \bar{X}$
  • B
    1
  • C
    $\Sigma f_1$
  • ✓
    $0$
Answer
Correct option: D.
$0$
(d)
We have
$\bar{X}=\frac{1}{N} \sum_{i=1}^n f_1 x_1 \Rightarrow \sum_{i=1}^n f_1 x_i=N \bar{X}$
$\sum_{i=1}^n f_1\left(x_i-\bar{X}\right)=\sum_{i=1}^n f_i x_i-\sum_{i=1}^n f_i \bar{X}=N \bar{X}-\bar{X} \sum_{i=1}^n f_i=N \bar{X}-N \bar{X}=0$
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M.C.Q (1 Marks) - Page 2 - MATHS STD 10 Questions - Vidyadip