MCQ 511 Mark
The most frequently occurring data value in a data set is the $...........$
AnswerMode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and,
which represents the whole series as most of the values in the series revolves around this value.
View full question & answer→MCQ 521 Mark
Choose the correct answer. Let $x_1, x_2, ...,x_n$ be n observations and $\bar{\text{x}}$ be their arithmetic mean. The formula for the standard deviation is given by:
- A
$\sum(\text{x}_\text{i}-\bar{\text{x}})^2$
- B
$\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{x}}$
- ✓
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
- D
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
AnswerCorrect option: C. $\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
The formula for $\text{S.D}=\sigma=\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
View full question & answer→MCQ 531 Mark
Choose the correct answer. The mean deviation of the data $3, 10, 10, 4, 7, 10, 5$ from the mean is:
AnswerCorrect option: B. $2.57$
Observations are fiven by $3, 10, 10, 4, 7, 10,$ and $5$
$\therefore\ \bar{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
$\text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{18}{7}=2.57$
|
$x_i$
|
$\text{d}_\text{i}=|\text{x}_\text{i}-\bar{\text{x}}|$
|
| $3$ |
$4$ |
| $10$ |
$3$ |
| $10$ |
$3$ |
| $4$ |
$3$ |
| $7$ |
$0$ |
| $10$ |
$3$ |
| $5$ |
$2$ |
|
Total
|
$\sum\text{d}_\text{i}=18$
|
View full question & answer→MCQ 541 Mark
The sum $\displaystyle \sum _{\text{ r}=1 }^{ 10 }{ \left( {\text{ r} }^{ 2 }+1 \right) \times \left( \text{r}\ ! \right) }$ is equal to:
- A
$(11)!$
- ✓
$10 \times (11)!$
- C
$101 \times (10)!$
- D
$101 \times (10)!$
AnswerCorrect option: B. $10 \times (11)!$
$ \sum(\text{r}^2+1)\text{r}!=\sum[\text{r}(\text{r }+1) -(\text{r}-1)\text{r}!$
$ =\sum\limits^{10}_\text{r=1}[\text{r}(\text{r }+1)!-(\text{r}-1)\text{r}!]$
$= (1\times 2!−0\times 1!)+(2\times 3!−1\times 2!)+......+(10\times 11!−9\times 10!)$
$=10\times 11! $
View full question & answer→MCQ 551 Mark
If the first and the second letters of the word $\text{MISJUDGEMENTS}$ are interchanged with the last and the second last letters respectively, and similarly the third and the fourth letters are interchanged with the third and the fourth letters from the last respectively , and so on,then what will be the fifth letter to the right of the third letter from the left end?
Answer
| $M$ |
$I$ |
$S$ |
$J$ |
$U$ |
$D$ |
$G$ |
$E$ |
$M$ |
$E$ |
$N$ |
$T$ |
$S$ |
| $1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
$9$ |
$10$ |
$11$ |
$12$ |
$13$ |
The alphabet series is shown in the above diagram with positions.
Now, with the required interchange in positions,
as the question says, we get following letters interchanged
$1^{st} \rightarrow 13^{th}$
$2^{nd} \rightarrow 12^{th}$
$3^{rd} \rightarrow 11^{th}$
$4^{th} \rightarrow 10^{th}$
$5^{th} \rightarrow 9^{th}$
$6^{th} \rightarrow 8^{th}$
$7^{th} \rightarrow 7^{th}$
Using above mentioned interchange we get complete reversal of the letters of this word.
which is: $\text{STNEMEGDUJSIM}$
Here Third letter from the left end is $N.$
According to the question,
$\Rightarrow (5 + 3)^{th}$ letter from the left
$\Rightarrow 8^{th}$ letter from the left Hence, $8^{th}$ letter from the left end is $D.$ View full question & answer→MCQ 561 Mark
The mean of $x, x + 3, x + 4, x + 8$ and $x + 10:$
- A
$x + 4$
- B
$x + 8$
- C
$x + 3$
- ✓
$x + 5$
AnswerCorrect option: D. $x + 5$
By definition
$\text{Average} =\cfrac{\text{x}+(\text{x}+3)+(\text{x}+4)+((\text{x}+8)+(\text{x}+10)}{5}$
$=\cfrac{5\text{x}+25}{5}$
$=(\text{x}+5)$
View full question & answer→MCQ 571 Mark
The average can be found only in $............$ variables:
AnswerThe average can be found only in quantitative variables.
Example: A quantitative variable is something that
can be measured and written out as a number.
So, we can find the average marks of $2020$ students in $1212$ class but,
we can not find the average of the
cleverness of students.
View full question & answer→MCQ 581 Mark
Choose the correct answer. Coefficient of variation of two distributions are $50$ and $60,$ and their arithmetic means are $30$ and $25$ respectively. Difference of their standard deviation is:
AnswerHere, $\text{CV}_1=50,\ \text{CV}_2=60,\ \bar{\text{x}}_1=30$ and $\bar{\text{x}}_2=25$
$\therefore\ \text{CV}_1=\frac{\sigma_1}{\bar{\text{x}}_1}\times100$
$\Rightarrow50=\frac{\sigma_1}{30}\times100$
$\therefore\ \sigma_1=\frac{30\times50}{100}=15$
and $\text{CV}_2=\frac{\sigma_2}{\bar{\text{x}}_2}\times100$
$\Rightarrow60=\frac{\sigma_2}{25}\times100$
$\therefore\ \sigma^2=\frac{60\times25}{100}=15$
Now, $\sigma_1-\sigma_2=15-15=0$
View full question & answer→MCQ 591 Mark
If the mode of five observations, in order, $0, 2, 3, m, 5$ is $3$ then $m = ...........$
AnswerIf the mode of five observation, in order, $0, 2, 3, m, 5$ is $3$, then $m=3.$
View full question & answer→MCQ 601 Mark
For dealing with qualitative data the best average is:
AnswerMedian is the middle most value.
Also for even number of observations, median is the average of to middle values.
Hence, it divides the whole series into two equal halv.
It gives the more accurate and best average for qualitative data.
View full question & answer→MCQ 611 Mark
If the arithmetic mean of $7, 8, x, 11, 14$ is $x,$ then $x:$
AnswerWe have,
$= \frac{7+8+\text{x}+11+14}{5}$
$= \text{x} = 40+\text{x} = 5\text{x} = \text{x} = 10$
View full question & answer→MCQ 621 Mark
The mean of $100$ observations is $50$ and their standard deviation is $5.$ The sum of all squares of all the observations is:
- A
$50,000$
- B
$250,000$
- ✓
$252500$
- D
$25000$
AnswerCorrect option: C. $252500$
Let $\overline{\text{x}}$ and $\sigma$ be the mean and standard deviation of $100$ observations, respectively.
$\therefore\overline{\text{x}}=50,\ \sigma=5$ and $n = 100$
$\text{Mean},\ \overline{\text{x}}=50$
$\Rightarrow\frac{\sum\text{x}_\text{i}}{100}=50$
$\Rightarrow\sum\text{x}_\text{i}=5000\ ...(1)$
Now,
Standard deviation, $\sigma=5$
$\Rightarrow\sqrt{\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{\sum\text{x}_\text{i}}{100}\Big)^2}=5$
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{5000}{100}\Big)^2=25 [$From $(1)]$
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}=25+2500=2525$
$\Rightarrow{\sum\text{x}_\text{i}^2}=252500$
Thus, the sum of all squares of all the observations is $252500.$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 631 Mark
The following data has been arranged in ascending order. If their median is $63,$ find the value of $x.34, 37, 53, 55, x, x + 2, 77, 83, 89$ and $100.$
AnswerThe series in ascending order is: $34, 37, 53, 55, x, x + 2, 77, 83, 89$ and
The series has $10$ numbers, even numbers.
Hence, the median will be the mean of the two middle numbers:
median $=$ mean of $5^{th}$ and $6^{th}$ terms
$63=\frac{\text{x}\ + \ \text{x}\ +\ 2}{2}$
$1260=2\text{x}+2$
$2\text{x}=124$
$\text{x}=62$
View full question & answer→MCQ 641 Mark
If $150$ is the mean of $200$ observations and $100$ is the mean of some $300$ other observations, find the mean of the combination:
AnswerMean of $200$ observations $= 150$
Sum of $200$ observations $= 200 \times 150 = 30000$
Mean of $300$ observations $= 100$
Sum of $300$ observations $= 300 \times 100 = 30000$
Total Sum $= 30000 + 30000 = 60000$
Number of observations $= 100 + 200 = 500$
$\text{Mean} = \frac{\text{Sum}}{\text{Number of observations}}$
$= \frac{60000}{500}$
$= 120$
View full question & answer→MCQ 651 Mark
if $x_1, x_2, x_3, x_4, x_5$ are five consecutive odd numbers, then their average is:
AnswerThe five consecutive odd numbers are $\text{x}_1+ \text{x}_1+2, \text{x}_1 + 4,\text{x}_1 +6,\text{x}_1 +8$
$\therefore \text{mean}=\frac{\text{x}_1 \ + \ \text{x}_1 \ +\ 2+\text{x}_1 \ +\ 4\ +\ \text{x}_1 \ +\ 6\text{x}_1 \ +\ 8}{6}$
$=\frac{5\text{x}_1\ +\ 20}{5}$
$=\text{x}_1\ +\ 4$
$=\text{x}_3$
View full question & answer→MCQ 661 Mark
The standard deviation of the observations $6, 5, 9, 13, 12, 8, 10$ is:
- A
$6$
- B
$\sqrt6$
- C
$\frac{52}{7}$
- ✓
$\sqrt{\frac{52}{7}}$
AnswerCorrect option: D. $\sqrt{\frac{52}{7}}$
The given observations are $6, 5, 9, 13, 12, 8, 10.$
Now,
$\sum\text{x}_\text{i}= 6 + 5 + 9 + 13 + 12 + 8 + 10 = 63$
$\sum\text{x}_\text{i}^2=36 + 25 + 81 + 169 + 144 + 64 + 100 = 619$
$\therefore$ Standard deviation of the observations, $\sigma$
$=\sqrt{\frac{1}{\text{N}}\sum\text{x}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{x}_\text{i}\Big)^2}$
$=\sqrt{\frac{1}{7}\times619-\Big(\frac{1}{7}\times63\Big)^2}$
$=\sqrt{\frac{619}{7}-81}$
$=\sqrt{\frac{619-567}{7}}$
$=\sqrt{\frac{52}{7}}$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 671 Mark
The average of a collection of $20$ measurements was calculated to be $56 \ cm.$ But later it was found that a mistake occured in one of the measurements which was recorded as $64 \ cm$ but should have been $61 \ cm.$ What is the correct average?
- A
$39.55\ cm$
- B
$29.55 \ cm$
- C
$55.85 \ cm$
- ✓
$55.85 \ cm$
AnswerCorrect option: D. $55.85 \ cm$
Incorrect total of $20$ measurement $= 20 \times 56 = 1120$
Correct total $= 1120 - 64 + 61 = 1117$
$\therefore \text{Correct average} = \displaystyle \frac{1117}{20} = 55.85\ \text{cm}$
View full question & answer→MCQ 681 Mark
If the mean of five observations $x ,x + 2, x + 4, x + 6$ and $x + 8$ is $11,$ then the mean of last three obsevations is:
AnswerGiven observations $x, x + 2, x + 4, x + 6, x + 8$
$\Rightarrow\frac{5\text{x}+ 20}{5}=11$
$\Rightarrow x = 7$
So, the observations are $7, 9, 11, 13, 15$
Req. mean $=\frac{11+13+15}{3}=13$
View full question & answer→MCQ 691 Mark
The standard deviation of the data:
| $x$ |
$1$ |
$a$ |
$a^2$ |
....
|
$a^n$ |
| $f$ |
${ }^n C_0$ |
${ }^n C_1$ |
${ }^n C_2$ |
....
|
${ }^n C_2$ |
is, - A
$\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
- B
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
- C
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
- ✓
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
AnswerCorrect option: D. $\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
|
$x_i$
|
$f_i$
|
$f_ix_i$
|
${x_i}^2$
|
${f_ix_i}^2$
|
|
1
|
${ }^n C_0$ |
${ }^n C_0$ |
1
|
1
|
|
a
|
${ }^n C_1$ |
$a{ }^n C_1$ |
$a^2$ |
$a^2{ ~}^n C_1$ |
|
a
|
${ }^n C_2$ |
$a^2{ ~}^n C_2$ |
$a^4$ |
$a^4{ ~}^n C_2$ |
|
a
|
${ }^n C_3$ |
$a^3{~}^n C_3$ |
$a^6$ |
$a^6{ ~}^n C_3$ |
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
|
$a^n$
|
${ }^n C_n$ |
$a^n{~ }^n C_n$ |
$a^{2n}$ |
$a^{2n}{~ }^n C_n$ |
|
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}=2^\text{n}$
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}=(1+\text{a})^\text{n}$
|
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$ |
Number of terms, $\text{N}=\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}=2^\text{n}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=^\text{n}\text{C}_0+\text{a}^\text{n}\text{C}_1+\text{a}^2{^\text{ n}\text{C}_2+...+\text{a}^\text{n}{^\text{ n}\text{C}_\text{n}}}=(1+\text{a})^\text{n}$
$\overline{\text{X}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$=\frac{(1+\text{a})^\text{n}}{2^\text{n}}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$
$\sigma^2=\text{Variance}(\text{X})=\frac{1}{\text{N}}\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2-\Bigg(\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Bigg)^2$
$=\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}-\Big[\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}\Big]^2$
$=\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}$
$\sigma=\sqrt{\text{Variance}(\text{X})}$
$=\sqrt{\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}}$ View full question & answer→MCQ 701 Mark
Of the three numbers, the first is twice the second, and the second is twice the third. The average of the reciprocal of the numbers is $\frac{7}{72}$. What are the three numbers?
- ✓
$24, 12, 6$
- B
$8, 4, 2$
- C
$12, 6, 3$
- D
$12, 6, 3$
AnswerCorrect option: A. $24, 12, 6$
Let the third number be $x$ Then Second number $= 2x$
and first number $= 4x$
Sum of the reciprocals of these $3$ numbers $ =\frac{1}{\text{4x}}+\frac{1}{2\text{x}}+\frac{1}{\text{x}}=\frac{1+2+4}{4\text{x}}$
$= \frac{7}{\text{4x}}$
Given, $\frac{7}{4\text{x}}=3\times \frac{7}{72}$
$=4\text{x}=24= \text{x}=6 $
Therefore, the three numbers are,
$4 \times 6, 2 \times 6, 6$
i.e. $24, 12, 6.$
View full question & answer→MCQ 711 Mark
The weight of a body, calculated as the average of seven different experiments is $53.735g.$The average of the first three experiments is $54.005g.$ The fourth was greater than the fifth by $0.0040.004 g$ and the average of sixth and seventh was $0.010g$ less than the average of the first three. Find the weight of the body in the fourth experiment.
- A
$52.071g$
- ✓
$53.072g$
- C
$51.450g$
- D
$51.450g$
AnswerCorrect option: B. $53.072g$
View full question & answer→MCQ 721 Mark
The mean of $100$ numbers is $45.$ The mean of the last $99$ numbers is $44.$ The first number is
AnswerThe mean of $100$ numbers is $45$
$\therefore$ Sum of all $100$ numbers $= 100 \times 45 = 4500$
The mean of last $99$ numbers is $44$
$\therefore$ Sum of all last $99$ numbers $= 99 \times 44 = 4356$
$\Rightarrow$ The first number $= 4500 - 4356 = 144$
View full question & answer→MCQ 731 Mark
Which of the following is not changed for the observations $31, 48, 50, 60, 25, 8, 3x, 26, 32? ($where $x$ lies between $10$ and $15):$
View full question & answer→MCQ 741 Mark
Find the mean of first six natural numbers:
AnswerFirst six natural numbers $= 1, 2, 3, 4, 5, 6$
$\text{ Mean} = \frac{\text{Sum}}{\text{Number of observations}} $
$\text{Mean} = \frac{1 + 2 + 3 + 4 + 5 +6}{6} = \dfrac{21}{6}$
$ =3.5$
View full question & answer→MCQ 751 Mark
On Thursday, $20$ of the $25$ students in a chemistry class took a test and their average $($arithmetic mean$)$ was $80.$ On Friday, the other $5$ students took the test and their average $($arithmetic mean$)$ was $90.$ What was the average for the entire class?
AnswerAverage $= \frac{20\left ( 80 \right )\ +\ 5\left ( 90 \right )}{25}$
$=\frac{1600\ +\ 450}{25}$
$=\frac{2050}{25}$
$=82$
View full question & answer→MCQ 761 Mark
The mean deviation of the numbers $3, 4, 5, 6, 7$ from the mean is:
Answer$\text{Mean}(\overline{\text{X}})=\frac{3+4+5+6+7}{5}$
$=\frac{25}{5}$
$=5$
Taking the absolute value of deviation of each term from the mean, we get:
$\text{MD}=\frac{|(3-5)|+|(4-5)|+|(5-5)|+|(6-5)|+|(7-5)|}{5}$
$=\frac{2+1+0+1+2}{5}$
$=\frac{6}{5}$
$=1.2$
View full question & answer→MCQ 771 Mark
When there are $2$ observations in the middle, median is calculated by $.........$
AnswerCorrect option: D. Both $(B)$ and $(C) $
Median is the middle most value of a series.
So when the series has odd number of elements then,
median can be calculated easily but when the series has even number of elements then,
The series has two middle values, so
median is calculated either by taking out the average of both the
value or the median is the $\frac{(\text{N}+1)}{ 2}^{th}$ element of the series.
View full question & answer→MCQ 781 Mark
The average age of a group of eight is same as it was $3$ years ago when a young member is substituted for an old member the incoming member is younger to the outgoing member by:
- A
$11$ years
- ✓
$24$ years
- C
$25$ years
- D
$28$ years
AnswerCorrect option: B. $24$ years
let presently the member be $ x_1, x_2, x_3....x_8 $
So the average age $ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}......1$
Now the average age of all the members $3$ years ago
$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$
if $x_1$ the younger member is replaced by the older member $y_1$ then,
$=\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}$
$=\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$
$\Rightarrow \text{x}1=\text{y}1+24$
$\Rightarrow \text{x}_1-\text{y}_1=24 $
View full question & answer→MCQ 791 Mark
The mean of $x, y, z$ is $y,$ then $x + z =:$
View full question & answer→MCQ 801 Mark
Choose the correct answer. Standard deviations for first $10$ natural numbers is:
- A
$5.5$
- B
$3.87$
- C
$2.17$
- ✓
$2.97$
AnswerCorrect option: D. $2.97$
We know that $SD$ of first $n$ natural numbers $\sqrt{\frac{\text{n}^2-1}{12}}$
Here, $\text{n}=10$
$\therefore\ \text{SD}=\sqrt{\frac{(10)^2-1}{12}}$
$=\sqrt{\frac{99}{12}} $
$=\sqrt{8.25}$
$=2.97$
View full question & answer→MCQ 811 Mark
Given the following data set, what is the value of median $(2\ 4\ 3\ 6\ 1\ 8\ 9\ 2\ 5\ 7).$
AnswerMedian is the middle most value of a series.
So when the series has odd number of elements then median can be calculated easily but,
when the series has even number of elements then the series has two middle values,
so median is calculated by taking out the average of both the value.
The given series is first arranged into ascending; $1, 2, 2, 3, 4, 5, 6, 7, 8, 9$
$\text{N} = 10$
$\text{median}= \frac{(10+1)}{2}$
$\text{th}$ term $= \frac{11}{2}$
$\text{th}$ term $= 5.5 $
$=\frac{( \text{value of 5th term }+ \text{value of $6^{th}$ term)}}{2}$
$= \frac{(4+5)}{2}$
$= \frac{9}{2}$
$= 4.5$
View full question & answer→MCQ 821 Mark
| Size |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
| Frequency |
$10$ |
$12$ |
$25$ |
$20$ |
$25$ |
$15$ |
$11$ |
- A
$2$
- B
$8$
- ✓
Both $4$ and $6$
- D
Both $6$ and $8$
AnswerCorrect option: C. Both $4$ and $6$
Mode is that observation which have highest frequency.
Since, both $4$ and $6$ have highest frequency
i.e. $25$ and $25,$ they are the mode of the given distribution.
Hence, option $(C)$ is correct.
View full question & answer→MCQ 831 Mark
Choose the correct answer. Let ${x}_1, {x}_2, {x}_3, {x}_4, {x}_5$ be the observations with mean $m$ and standard deviation $s$. The standard deviation of the observations $k x_1, k x_2, k x_3, k x_4, k x_5$ is:
AnswerHere, $\text{m}=\frac{\sum\text{x}_\text{i}}{\text{N}},$
$\text{S}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$\therefore\ \text{SD}=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\text{K}\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\text{K}^2\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}.\text{S}$
View full question & answer→MCQ 841 Mark
Mean of a set of $23$ values is $7$, if each value is multiplied by $23$ the new mean is:
Answer$\Rightarrow$ Sum of the observation $= 23 \times 7 = 161$
If each observation is multiplied by $23$ then the sum is also multiplied by $23.$
New sum $= 23 \times 161 = 3703$
$\Rightarrow$ New mean $ =\frac{3703}{23}=161$
View full question & answer→MCQ 851 Mark
A batsman scores runs in $10$ innings as $38, 70, 48, 34, 42, 55, 63, 46, 54$ and $44.$ The mean deviation about mean is:
- ✓
$8.6$
- B
$6.4$
- C
$10.6$
- D
$10.6$
Answer$N = 10$
$\overline{\text{X}}=\frac{38+70+48+34+42+55+63+46+54+44}{10}$
$=\frac{494}{10}$
$=49.4$
|
$x_i$
|
$\text{d}_\text{i}=\big|\text{x}_\text{i}-49.4\big|$
|
| $34$ |
$15.4$ |
| $38$ |
$11.4$ |
| $42$ |
$7.4$ |
| $44$ |
$5.4$ |
| $46$ |
$3.4$ |
| $48$ |
$1.4$ |
| $54$ |
$4.6$ |
| $55$ |
$5.6$ |
| $63$ |
$13.6$ |
| $70$ |
$20.6$ |
|
|
$\sum\limits^{\text{n}}_{\text{i}=}\text{d}_\text{i}=88.8$
|
Mean deviation from the mean $=\frac{88.8}{10}= 8.88$
Disclaimer: No option is matching the answer. View full question & answer→MCQ 861 Mark
Median divides the total frequency into $..........$ equal parts:
AnswerThe median of the data series is the middle term or the mean of the two middle terms.
Hence, it divides the data series or the frequency of terms into two equal halves.
View full question & answer→MCQ 871 Mark
The average of $\displaystyle 1\frac{1}{6},2\frac{1}{3},6\frac{2}{3}161,231,632$ and $\displaystyle 8\frac{5}{6}865$ is:
- A
$\displaystyle 6\frac{3}{4}$
- B
$\displaystyle 5\frac{3}{4}$
- ✓
$\displaystyle 4\frac{3}{4}$
- D
$\displaystyle 4\frac{3}{4}$
AnswerCorrect option: C. $\displaystyle 4\frac{3}{4}$
$=\displaystyle 1\frac{1}{6}+2\frac{1}{3}+6\frac{2}{3}+ 8\frac{5}{6}$
$=\frac{7}{6}+ \frac{7}{3 }+ \frac{20}{3 }+\frac{53}{6}$
$ = \frac{6+7+14+40+53}{6} $
$= \frac{114}{6}$
$=19$
$\therefore \text{Average}=\frac{19}{4}=4\dfrac{3}{4}$
View full question & answer→MCQ 881 Mark
The average age of a teacher and three students is $20$ years. If all students are of equal age and the difference between the age of the teacher and that of a student is $20$ years, then the age of the teacher is:
- A
$25$ years
- B
$30$ years
- ✓
$35$ years
- D
$36$ years
AnswerCorrect option: C. $35$ years
Let the age of each student be $x$ years
Then, the age of teacher will be $(x + 20)$ years
Mean age $=\frac{\left (\text{x}+20 \right )+3\text{x}}{4}$
$20=\frac{\text{4x}+20}{4}$
$\Rightarrow x = 15$
Hence, age of the teacher $= 35$ years
View full question & answer→MCQ 891 Mark
The average of four consecutive even numbers is one fourth of the sum of these numbers. What is the difference between the first and last number?
AnswerLet the numbers be $2x - 2, 2x, 2x + 2$ and $2x + 4,$
where $x$ is a natural number.
Then the difference between the first and last number $= 2x + 4 - (2x - 2) = 6.$
View full question & answer→MCQ 901 Mark
The most frequent value in a data set is?
AnswerMode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and, which represents the whole series as most of the values, in the series revolves around this value.
Therefore, mode is the value that occurs the most frequent times in a series.
View full question & answer→MCQ 911 Mark
Consider the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$ If $1$ is added to each number, the variance of the numbers so obtained is:
- ✓
$8.25$
- B
$2.87$
- C
$3.87$
- D
$3.87$
AnswerCorrect option: A. $8.25$
The given numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$
If $1$ is added to each number, then the new numbers obtained are
$2, 3, 4, 5, 6, 7, 8, 9, 10, 11$
Now,
$\sum\text{x}_\text{i}= 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 65$
$\sum\text{x}_\text{i}^2= 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 = 505$
$\therefore$ Variance of the numbers so obtained
$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-42.25$
$=8.25$
View full question & answer→MCQ 921 Mark
A child says that the median of $3, 14, 18, 20, 5$ is $18.$ What concept does the child missed about finding the median?
AnswerTo calculate the median of any data series.
The data series has to be arranged in the ascending order.
The child hasn't arranged the data series in ascending order.
View full question & answer→MCQ 931 Mark
The captain of a cricket team of $11$ members is $26$ years old and the wicket keeper is $3$ years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?
- ✓
$23$ years
- B
$24$ years
- C
$25$ years
- D
$25$ years
AnswerCorrect option: A. $23$ years
Let the average age of the whole team by $x$ years.
$= 11 x - (26 + 29) = 9(x - 1)$
$= 11x - 9x = 46$
$= 2x = 46$
$\Rightarrow 2x = 46$
$= x = 23$
$\Rightarrow x = 23.$
So, average age of the team is $23$ years.
View full question & answer→MCQ 941 Mark
Two high school classes took the same test. One class of $20$ students made an average grade of $80\%;$ the other class of $30$ students made an average grade of $70\%.$ The average grade for all students in both classes is:
- A
$75\%$
- ✓
$74\%$
- C
$77\%$
- D
$77\%$
AnswerCorrect option: B. $74\%$
$\text{Average}=\frac{20.80+30.70}{20+30}=74$
View full question & answer→MCQ 951 Mark
A school has $20$ teachers one of them retires at the age of $60$ years and a new teacher replaces him this change reduces the average age of the staff by $2$ years the age of new teacher is:
- A
$28$ years
- B
$25$ years
- ✓
$20$ years
- D
$21$ years
AnswerCorrect option: C. $20$ years
Let the average age of the staff $= x$
Age of the new teacher$= y$
According to the questionNew age of the staff reduced by
$2$ years $\Rightarrow \dfrac{20\text{x}-60+\text{y}}{20}$
$\text{ x}-2\Rightarrow 20\text{x}−60+\text{y}$
$\Rightarrow 20\text{x}-60+\text{y}$
$ =20(\text{x}-2)$
$\Rightarrow 20\text{x}−60+\text{y}$
$\Rightarrow 20\text{x}−60+\text{y}$
$\Rightarrow\text{y}=60-40=20$
$\Rightarrow \text{y}=60−40= 20$
Hence the age of the new teacher is $20$ years.
View full question & answer→MCQ 961 Mark
For a frequency distribution standard deviation is computed by applying the formula:
- ✓
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
- B
$\sigma=\sqrt{\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2-\frac{\sum\text{fd}^2}{\sum\text{f}}}$
- C
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\frac{\sum\text{fd}}{\sum\text{f}}}$
- D
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\frac{\sum\text{fd}}{\sum\text{f}}}$
AnswerCorrect option: A. $\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
View full question & answer→MCQ 971 Mark
Choose the correct answer. Following are the marks obtained by $9$ students in a mathematics test: $50, 69, 20, 33, 53, 39, 40, 65, 59$ The mean deviation from the median is:
- A
$9$
- B
$10.5$
- ✓
$12.67$
- D
$10.67$
AnswerCorrect option: C. $12.67$
$\therefore\ \text{Median}=5^{\text{th}}\text{ term}$
$\text{M}_\text{e}=50$
| $x_i$ |
$d_i = |x_i - M_e|$ |
| $20$ |
$30$ |
| $33$ |
$17$ |
| $39$ |
$11$ |
| $40$ |
$10$ |
| $50$ |
$0$ |
| $53$ |
$3$ |
| $59$ |
$9$ |
| $65$ |
$15$ |
| $69$ |
$19$ |
| $N = 2$ |
$\sum\text{d}_\text{i}=114$ |
$\therefore\ \text{MD}=\frac{114}{9}=12.67$ View full question & answer→MCQ 981 Mark
AnswerMode is the value that occurs most often For example:
$13, 13, 12, 14, 13$ The Mode of the following is $13.$
View full question & answer→MCQ 991 Mark
The mean of $20$ observations is $15$ On checking it was found that the two observations were wrongly copied as $3$ and $6.$ The correct values are $8$ and $4$ , then correct mean will be given by:
- ✓
$15.15$
- B
$14.69$
- C
$14.74$
- D
$15.54$
AnswerCorrect option: A. $15.15$
Mean of $20$ observatios $= 15$
Sum of $20$ observations $= 15 \times 20 = 300$
Correct sum $= 300 + 8 + 4 - 3 - 6 = 300$
Correct mean $= \dfrac{303}{20} = 15.15$
View full question & answer→MCQ 1001 Mark
The mean of $6$ numbers is $42$ If one number is excluded, the mean of remaining numbers is $45.$ Find the excluded number:
Answermean of $6$ numbers $= 42$
Sum of $6$ numbers $= 42 \times 6 = 252$
After excluding one number,
mean of $5$ numbers $= 45$
Sum of $5$ numbers $= 45 \times 5 = 225$
Thus, the number excluded $= 252 - 225 = 27$
View full question & answer→