MCQ 11 Mark
AnswerMost Frequent value is called mode.
View full question & answer→MCQ 21 Mark
The algebraic sum of the deviations of a set of n values from their mean is:
Answerif is the mean of $n$ observations $x_1, x_2, x_3, x_4 \ldots x_n$.
then algebraic sum of deviations $=\sum\limits^\text{n}_{\text{i}=0}\Big(\text{x}_\text{i}-{\overline{\text{X}}}\Big)$
$=\sum\limits^\text{n}_{\text{i}=0}\text{x}_\text{i}-\text{n}{\overline{\text{X}}}$
$=\text{n}\bigg(\frac{\sum^\text{n}_{\text{i}=0}\text{x}_\text{i}}{\text{n}}\bigg)-\text{n}\overline{\text{X}}$
$=\text{n}\overline{\text{X}}-\text{n}\overline{\text{X}}$
$=0$
View full question & answer→MCQ 31 Mark
For the set of numbers $2, 2, 4, 5$ and $12$, which of the following statements is true?
AnswerMedian $= 4$
Mode $= 2$
$\text{Mean}=\frac{2+2+4+5+12}{5}=\frac{25}{5}=5$
Hence, (Mean = $5$) > (Mode = $2$)
View full question & answer→MCQ 41 Mark
The mean of n observations is $\overline{\text{X}}.$ If each observation is multiplied by $k$, the mean of new observations is:
- ✓
$\text{k}\overline{\text{X}}$
- B
$\frac{\overline{\text{X}}}{\text{k}}$
- C
$\overline{\text{X}}+\text{k}$
- D
$\overline{\text{X}}-\text{k}$
AnswerCorrect option: A. $\text{k}\overline{\text{X}}$
$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{\text{Sum of all observations}}{\text{n}}$
if each observation is multiplied by $k$, then
$\text{New Mean},\overline{\text{X'}}=\frac{(\text{Sum of all observations})\text{k}}{\text{n}}$
$\Rightarrow\overline{\text{X}'}=\text{k}\overline{\text{X}}$
View full question & answer→MCQ 51 Mark
The empirical relation between mean, mode and median is:
- ✓
Mode = $3$ Median - $2$ Mean.
- B
Mode = $2$ Median - $3$ Mean.
- C
Median = $3$ Mode - $2$ Mean.
- D
Mean = $3$ Median - $2$ Mode.
AnswerCorrect option: A. Mode = $3$ Median - $2$ Mean.
The empirical Relation between mean, median and mode is:
Mode = $3$ Median - $2$ mean.
View full question & answer→MCQ 61 Mark
If the mean of five observations $x, x + 2, x + 4, x + 6, x + 8,$ is $11$, then the mean of first three observations is:
AnswerMean of first five observations $=\frac{\text{x}+\text{x}+2+\text{x}+4+\text{x}+6+\text{x}+8}{5}=11$
$\Rightarrow 5x + 20 = 55$
$\Rightarrow x = 7$
$\Rightarrow $ First three numbers are $7, 9, 11$
$\text{Mean}=\frac{7+9+11}{3}=\frac{27}{3}=9$
View full question & answer→MCQ 71 Mark
The mean of $a, b, c, d$ and e is $28$. If the mean of $a, c,$ and $e$ is $24$, What is the mean of $b$ and $d$?
Answer$\text{Mean}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}=28$
$\Rightarrow a + b + c + d + e = 140 ...(1)$
Also, $\text{Mean}=\frac{\text{a}+\text{c}+\text{e}}{3}=24$
$\Rightarrow a+ c + e = 72 ...(2)$
Subtracting equation $(2)$ from $(1)$, we have
$b + d = 68$
$\text{Mean}=\frac{\text{b}+\text{d}}{2}=\frac{68}{2}=34$
View full question & answer→MCQ 81 Mark
The following is the data of wages per day : $5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8$ The mode of the data is:
AnswerIn data $5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8,$
We observe that values $5$ and $8$ both have maximum frequency i.e. $4$
View full question & answer→MCQ 91 Mark
If the arithmetic mean of $7, 5, 13, x$ and $9$ is $10$, then the value $x$ is:
Answer$\text{Mean}=\frac{7+5+13+\text{x}+9}{5}=10$
$\Rightarrow34+\text{x}=50$
$\Rightarrow\text{x}=16$
View full question & answer→MCQ 101 Mark
The median of the following data : $0, 2, 2, 2, -3, 5, -1, 5, −3, 6, 6, 5, 6$ is:
AnswerData: $0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6$
Rearranging data in increasing order, we have
$-3, -3, -1, 0, 2, 2, 5, 5, 5, 5, 6, 6, 6$
Number of observations $= n = 14$ (even)
Now,
$\text{Median}=\frac{\Big(\frac{\text{n}}{2}\Big)^{\text{th}}\text{observation}+\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}}{2}$
$=\frac{7^{\text{th}}\text{observation}+8^{\text{th}}\text{observation}}{2}$
$=\frac{2+5}{2}$
$\Rightarrow\text{Median}=3.5$
View full question & answer→MCQ 111 Mark
Which one of the following is not a measure of central value?
AnswerThe difference between the highest value and the lowest value in the data set is called Range.
View full question & answer→MCQ 121 Mark
$A, B, C$ are three sets of values of $x$:
$A: 2, 3, 7, 1, 3, 2, 3$
$B: 7, 5, 9, 12, 5, 3, 8$
$C: 4, 4, 11, 7, 2, 3, 4$
Which one of the following statements is correct? - A
Mean of $A$ = Mode of $C$
- B
Mean of $C$ = Median of $B$
- C
Median of $B$ = Mode of $A$
- ✓
Mean, Median and Mode of $A$ are equal.
AnswerCorrect option: D. Mean, Median and Mode of $A$ are equal.
$A: 1, 2, 2, 3, 3, 3, 7$
$B: 3, 5, 5, 7, 8, 9, 12$
$C: 2, 3, 4, 4, 4, 7, 11$
$\text{Mean of A}=\frac{1+2+2+3+3+3+7}{7}$
$=\frac{21}{7}=3$
$\text{Mean of B}=\frac{3+5+5+7+8+9+12}{7}$
$=\frac{49}{7}=7$
$\text{Mean of C}=\frac{2+3+4+4+4+7+11}{7}$
$=\frac{35}{7}=5$
Mode of $A = 3$; Median of $A = 3$
Mode of $B = 5$; Median of $B = 7$
Mode of $C = 4$; Median of $C = 4$
(Mean of $A = 3$) $\neq$ (Mode of $C = 4$)
(Mean of $C = 5$) $\neq$ (Median of $B = 4$)
(Median of $B = 7$) $\neq$ (Mode of $A = 3$)
Mean of $A = 3,$ Mode of $A = 3$, Median of $A = 3$ View full question & answer→MCQ 131 Mark
The mean of n observations is $\overline{\text{X}}.$ If $k$ is added to each observation, then the new mean is:
- A
$\overline{\text{X}}$
- ✓
$\overline{\text{X}}+\text{k}$
- C
$\overline{\text{X}}-\text{k}$
- D
$\text{k}\overline{\text{X}}$
AnswerCorrect option: B. $\overline{\text{X}}+\text{k}$
$\text{Mean}=\overline{\text{X}}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
$=\frac{\text{Sum of all observations}}{\text{n}}$
Now if $k$ is aded to each observation
$\text{New Mean},\overline{\text{X'}}=\frac{\text{Sum of all observations}+\text{nk}}{\text{n}}$
$=\frac{\text{Sum of all observations}}{\text{n}}+\text{k}$
$\Rightarrow\overline{\text{X}'}=\overline{\text{X}}+\text{k}$
View full question & answer→MCQ 141 Mark
The mean of a set of seven numbers is $81$. If one of the numbers is discarded, the mean of the remaining numbers is $78$. The value of discarded number is:
Answer$\text{Mean}=81=\frac{\text{Sum of seven numbers}}{7}$
$\Rightarrow $ Sum of seven numbers $= 81 \times 7 = 567$
Let the discared number be $x$.
$\Rightarrow $ Sum of $6$ numbers $= 567 - x$
Now, mean of remaining $6$ numbers $=\frac{567-\text{x}}{6}=78$
$\Rightarrow 567 - x = 468$
$\Rightarrow x = 99$
So, discarded number is $99$
View full question & answer→MCQ 151 Mark
For which set of numbers do the mean, median and mode all have the same value?
- A
$2, 2, 2, 2, 4$
- ✓
$1, 3, 3, 3, 5$
- C
$1, 1, 2, 5, 6$
- D
$1, 1, 1, 2, 5$
AnswerCorrect option: B. $1, 3, 3, 3, 5$
|
|
Mean
|
Median
|
Mode
|
|
$2, 2, 2, 2 4$
|
$\frac{12}{5}=2.4$
|
$2$
|
$2$
|
|
$1, 3, 3, 3, 5$
|
$\frac{15}{5}=3$
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$3$
|
$3$
|
|
$1, 1, 2, 5, 6$
|
$\frac{15}{5}=3$
|
$2$
|
$1$
|
|
$1, 1, 1, 2, 5$
|
$\frac{10}{5}=2$
|
$1$
|
$1$
|
From above table, data $1, 3, 3, 3, 5$ has mean, median, mode all have same value, i.e. $3$ View full question & answer→