M.C.Q (1 Marks)

MCQ 1511 Mark

The average weight of a group of $20$ boys was calculated to be $89.4\ kg$ and it was later discovered that one weight was misread as $78\ kg$ instead of $87\ kg.$ The correct average weight is:

A
$88.95\ kg$
✓
$89.85\ kg$
C
$89.55\ kg$
D
$87.55\ kg$

Answer

Correct option: B.

$89.85\ kg$

Difference in weight $= 87 - 78 = 9\ kg$
$\therefore$ Correct average weight $= 89.4 +\frac {9}{20}$
$= 89.4 + 0.45$
$= 89.85\ kg$

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MCQ 1521 Mark

Find the incorrect formula from the following:

A
$ \text{mode}=\text{L}+\Big(\frac{\text{f}\text{m}-\text{f}1}{\text{f}\text{m}-\text{f}1-\text{f}2}\Big)\text{h}$
B
$\text{Mean} = \frac{\sum {\text{ fixi }} }{\sum { \text{xi }} }$
✓
$\overline{\text{x}}= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }$
D
$\overline{\text{x}}= \frac{\sum {\text{ fi }} }{\sum { \text{fixi }} }$

Answer

Correct option: C.

$\overline{\text{x}}= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }$

$\text{ A Median} = \text{L} + \begin{pmatrix} \frac{N}{2} - \text{c.f.} \end{pmatrix} \frac{\text{h}}{\text{f}} , $
$ \text{mode}=\text{L}+\Big(\frac{\text{f}\text{m}-\text{f}1}{\text{f}\text{m}-\text{f}1-\text{f}2}\Big)\text{h}$ Where,
$L$ is lower boundary of median class,
$N$ is sum of frequencies, $c.f.$ is cumulative frequency,
$h$ is width of classes, $f$ is frequency of mode class,
$\text{f}_\text{m}$ is frequency of modal class,
$\text{f}_\text{1}$ is frequency of pre$-$modal class,
$\text{f}_\text{2}$ is frequency of post$-$modal class Mean is
given $= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }$
Thus, Mean given in option $C$ is incorrect.

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MCQ 1531 Mark

The algebraic sum of the deviations of a set of value of a data from their mean is:

A
$>0$
✓
$0$
C
$<1$
D
$<0$

Answer

Correct option: B.

$0$

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MCQ 1541 Mark

The mean deviation from the median is:

A
Equal to that measured from another value.
B
Maximum if all observations are positive.
C
Greater than that measured from any other value.
✓
Less than that measured from any other value.

Answer

Correct option: D.

Less than that measured from any other value.

In a frequency distribution, the sum of absolute values of deviations from the mean and mode is always more than the sum of the deviations from the median.

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MCQ 1551 Mark

Consider the first $10$ positive integers. If we multiply each number by $-1$ and then add $1$ to each number, the variance of the numbers so obtained is:

✓
$8.25$
B
$6.5$
C
$3.87$
D
$4.87$

Answer

Correct option: A.

$8.25$

The first $10$ positive integers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$
Multiplying each number by $−1,$ we get
$-1, -2, -3, -4, -5, -6, -7, -8, -9, -10$
Adding $1$ to each of these numbers, we get
$0, -1, -2, -3, -4, -5, -6, -7, -8, -9$
Now,
$\sum\text{x}_\text{i}= 0 + (-1) + (-2) + (-3) + (-4) + (-5) + (-6) + (-7) + (-8) + (-9) = -45$
$\sum\text{x}_\text{i}^2= 0 + 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 = 285$
$\therefore$ Variance of the obtained numbers
$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2$
$=28.5-20.25$
$=8.25$

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MCQ 1561 Mark

Calculate the mode for $17, 12, 19, 11, 20, 11, 20, 19, 10, 25, 19:$

A
$12$
B
$17$
✓
$19$
D
$21$

Answer

Correct option: C.

$19$

Mode is the value which occurs most often in the data set of values.
Given data set is $17, 12, 19, 11, 20, 11, 20, 19, 10, 25, 19$
In the above data set, value $19$ has occurred many times i.e., $3$ times.
Therefore the mode of the given data set is $19$

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MCQ 1571 Mark

The average weight of $20$ students was calculated $70\ kg.$ It was later discovered that one weight was misread as $70$ instead of $90,$ the correct average in $kg$ is

A
$80$
B
$72$
C
$75$
✓
$71$

Answer

Correct option: D.

$71$

Wrong average $=70$
$\therefore$ Wrong sum of weight of $20$ students $= 20 \times 70$
$ \therefore$ Correct sum of weights of $20$ students
$= 1400 -$ wrong weight $+$ correct weight
$= 1400 - 70 + 90$
$= 1420$
$\therefore \text{Correct mean} = \frac{1420}{20}=71 \text{kg.}$

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MCQ 1581 Mark

The sum of $12$ observations is $600$ then their mean is $............$

A
$30$
B
$40$
✓
$50$
D
$60$

Answer

Correct option: C.

$50$

$\text{Mean} = \frac { \text{Sum of observations}}{\text{Total number of observations}} = \dfrac {600}{12} = 50$

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MCQ 1591 Mark

Mean of twenty observations is $15.$ It two observations $3$ and $14$ are replaced by $8$ and $9$ respectively, then the new mean will be:

A
$14$
✓
$15$
C
$16$
D
$17$

Answer

Correct option: B.

$15$

Total $= 20 \times 15 = 300$
When observations $3$ and $14$ are replaced by $8$ and $9$ respectively,
new sum will be $300 - 3 - 14 + 8 + 9 = 300$
New mean will be again $15$

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MCQ 1601 Mark

Let the observations at hand be arranged in increasing order. Which one of the following measures will not be affected when the smallest and the largest observations are removed?

A
Mean
✓
Median
C
Made
D
Mode

Answer

Correct option: B.

Median

If largest and smallest are removed the centre will remain intact.
Hence median will not be disturbed.

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MCQ 1611 Mark

The value of $ \displaystyle\sum _{ \text{p,q,r} }^{ }{ {\text{ p}}^{ 2 } } - \displaystyle\sum _{ \text{p,q,r} }^{ }{ {\text{q}}^{2}}$ is:

A
$ { \text{p} }^{ 2 }+{ \text{q} }^{ 2 }+{ \text{r} }^{ 2 }$
✓
$ 0$
C
$ 2{ \text{p} }^{ 2 }+2{ \text{q} }^{ 2 }-2{ \text{r} }$
D
$ 2{ \text{p} }^{ 2 }+2{ \text{q} }^{ 2 }+2{ \text{r} }$

Answer

Correct option: B.

$ 0$

Both the summations runs over $ \text{p,q,r}$
$\therefore \displaystyle\sum _ { \text{p,q,r }}^{ }{ { \text{p} }^{ 2 } } - \displaystyle\sum _ { \text{p,q,r }}^{ }{ { \text{p} }^{ 2 } } - (\text{p}^2 +\text{q}^2+\text{r}^2) - (\text{p}^2 +\text{q}^2+\text{r}^2)= 0$
Hence, the answer is $0.$

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MCQ 1621 Mark

If $v$ is the variance and $\sigma$ is the standard deviation, then:

A
$\text{v}=\frac{1}{\sigma^2}$
B
$\text{v}=\frac{1}{\sigma}$
✓
$\text{V}=\sigma^2$
D
$\text{V}=\sigma$

Answer

Correct option: C.

$\text{V}=\sigma^2$

The variance is the square of the standard deviation.

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MCQ 1631 Mark

The mean of three numbers is $10.$ The mean of other four numbers is $12.$ Find the mean of all the numbers:

A
$13.5$
✓
$11.15$
C
$14.15$
D
$12.15$

Answer

Correct option: B.

$11.15$

Mean of $3$ nos $= 10$ Total of $3$ nos is $10 \times 3 = 30$ Mean of other $4$ nos is $12$
Total of $4$ nos is $12 \times 4 = 48$
total of $4 + 3 = 7$
numbers is $48 + 30 = 78$
Mean of $7$ numbers is $ \cfrac{78}{7} = 11.15.$

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MCQ 1641 Mark

Find the average of all the number between $6$ and $34$ which are divisible by $5.$

A
$18$
✓
$20$
C
$24$
D
$22$

Answer

Correct option: B.

$20$

Numbers between $6$ and $34$ divisible by $5$ are $10, 15, 20, 25, 30.$
Required average $= \frac{10+15+20+25+30}{5}=\frac {100}{5}= 20$

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MCQ 1651 Mark

If the $S.D.$ of a set of observations is $8$ and if each observation is divided by $−2,$ the $S.D.$ of the new set of observations will be:

A
$-4$
B
$-8$
C
$8$
✓
$4$

Answer

Correct option: D.

$4$

If a set of observations, with $SD \sigma \sigma$ , are multiplied with a non$-$zero real number a, then $SD$ of the new observations will be $|\text{a}|\sigma.$
Dividing the set of observations by $-2$ is same as multiplying the observations by $\frac{1}{-2}.$
New $\text{S.D.}=\Big|-\frac{1}{2}\Big|\times8$
$=\frac{8}{2}$
$=4$

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MCQ 1661 Mark

The mean deviation for $n$ observations $x_1, x_2, ...,x_n$ from their mean $\overline{\text{X}}$ is given by:

A
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
✓
$\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
C
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
D
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$

Answer

Correct option: B.

$\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$

The mean deviation for $n$ observations $x_1, x_2, ...,x_n$ from their mean $\overline{\text{X}}$ is $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$
Disclaimer: There is some printing error in option $(b)$ given in the question.
The answer would be option $(b)$ if it given as $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$

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MATHS M.C.Q (1 Marks)