MCQ 11 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ Consider the following data
| $x_i$ |
$5$ |
$10$ |
$15$ |
$20$ |
$25$ |
| $f_i$ |
$7$ |
$4$ |
$6$ |
$3$ |
$5$ |
Then, the mean deviation about the mean is $6.32.$
Reason $(R)$ Consider the following data.
| $x_i$ |
$10$ |
$30$ |
$50$ |
$70$ |
$90$ |
| $f_i$ |
$4$ |
$24$ |
$28$ |
$16$ |
$8$ |
Then, the mean deviation about the mean is $15.$ - A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
|
$\text{x}_\text{i}$
|
$\text{f}_{\text{i}}$
|
$\text{f}_\text{i}\text{x}_\text{i}$
|
$|\text{x}_{\text{i}}-\bar{\text{x}}|$
|
$\text{f}_\text{i}|\text{x}_\text{i}-\bar{\text{x}}|$
|
|
$5$
|
$7$
|
$35$
|
$| 5 - 14 | = 9$
|
$63$
|
|
$10$
|
$4$
|
$40$
|
$|10 -14 | = 4$
|
$16$
|
|
$15$
|
$6$
|
$90$
|
$| 15 - 14 | = 1$
|
$06$
|
|
$20$
|
$3$
|
$60$
|
$|20 -14 | = 6$
|
$18$
|
|
$25$
|
$5$
|
$125$
|
$|25 - 14 | = 11$
|
$55$
|
|
Total
|
$\sum\text{f}_\text{i}=25$
|
$350$
|
|
$158$
|
Mean $\text{(x)}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{350}{25}=14$
$\therefore$ Mean deviation about mean
$\frac{\sum\text{f}_\text{i}|\text{x}_\text{t}-\bar{\text{x}}|}{\sum\text{f}_\text{i}}=\frac{158}{25}=6.32$
Reason
|
$\text{x}_\text{i}$
|
$\text{f}_{\text{i}}$
|
$\text{f}_\text{i}\text{x}_\text{i}$
|
$|\text{x}_{\text{i}}-\bar{\text{x}}|$
|
$\text{f}_\text{i}|\text{x}_\text{i}-\bar{\text{x}}|$
|
|
$10$
|
$4$
|
$40$
|
$| 10 - 50 | = 40$
|
$160$
|
|
$30$
|
$24$
|
$720$
|
$| 30 - 50 | = 20$
|
$480$
|
|
$50$
|
$28$
|
$1400$
|
$| 50 - 50 | = 00$
|
$000$
|
|
$70$
|
$16$
|
$1120$
|
$| 70 - 50 | = 20$
|
$320$
|
|
$90$
|
$8$
|
$720$
|
$| 90 - 50 | = 40$
|
$320$
|
|
Total
|
$\sum\text{f}_\text{i}=80$
|
$\sum\text{f}_\text{i}\text{x}_\text{i}=4000$
|
|
$1280$
|
Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{4000}{80}=50$
$\therefore$ Mean deviation about mean
$=\frac{\sum\text{f}_\text{i}|\text{x}_\text{i}-\bar{\text{x}}|}{\sum\text{f}_\text{i}}$
$=\frac{1280}{80}=16$ View full question & answer→MCQ 21 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The average marks of boys in a class is $52$ and that of girls is $42.$ The average marks of boys and girls combined is $50.$ The percentage of boys in the class is $80\%.$
Reason $(R)$ Mean marks scored by the students of a class is $53.$ The mean marks of the girls is $55$ and the mean marks of the boys is $50.$ The percentage of girls in the class is $64\%.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
Assertion Let the number of boys and girls be $x$ and $y.$
$\therefore 52x + 42y = 50(x + y)$
$\Rightarrow 2x = 8y$
$\Rightarrow x = 4y$
$\therefore$ Total number of students in the class $= x + y = 5y$
$\therefore$ Required percentage of boys
$=\frac{4\text{y}}{5\text{y}}\times100\%=80\%$
Reason Let the number of boys be $x$ and number of girls be $y.$
$\therefore 53(x + y) = 55y + 50x$
$\Rightarrow 3x = 2y$
$\Rightarrow\text{x}=\frac{2\text{y}}{3}$
$\therefore$ Total number of students
$=\text{x}+\text{y}=\frac{\text{2y}}{3}+\text{y}=\frac{5}{3}\text{y}$
Hence, required percentage
$=\frac{\text{y}}{\frac{5\text{y}}{3}}\times100\%$
$=\frac{3}{5}\times100\%=60\%$
View full question & answer→MCQ 31 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ Tf the mean of $n$ bservations $1^2, 2^2, 3^2, ..., n^2 $ is $\frac{46\text{x}}{11},$ then nis equal to $11.$
Reason $(R)$ For two data sets each of size $5,$ the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4,$ respectively. The variance of combined data set is $\frac{11}{2}.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A$.
- ✓
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: B. $A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
Assertion Mean of $1^2, 2^2, 3^2,...,n^2 $ is
$\frac{1^2+2^2+3^2+...+\text{n}^2}{\text{n}}=\frac{\sum\text{n}^2}{\text{n}}$
$\therefore\frac{46\text{n}}{11}\frac{\text{n}\text{(n+1}(2\text{n}+1)}{6\text{n}}$
$\Rightarrow 2\text{m} + 33\text{n}+11-276\text{n}=0$
$ \text{(n-11)} (22\text{n} -1)=0$
$\Rightarrow\text{n}=1\text{ and}\text{ n}\neq\frac{1}{22}$
Reason $\because\sigma_\text{x}^2=4 \text{ and } \sigma_\text{y}^2=5$
Also, $\bar{\text{x}}=2 \text{ and }\bar{\text{y}}=4$
Now, $\frac{\sum\text{x}_\text{i}}{5}=2\Rightarrow\sum\text{x}_\text{i}=10 \frac{\sum\text{y}_\text{i}}{5}=4$
$\Rightarrow\sum\text{y}_\text{i}=20$
Since, $\sigma_\text{x}^2=\frac{1}{5}(\sum\text{x}_\text{i}^2)-(\bar{\text{x}})^2$
$\Rightarrow\sum\text{x}_\text{i}^2=40$
Similarly, $\sum\text{y}_\text{i}^2=105$
$\because\sigma_\text{z}^2 = \frac{1}{10}(\sum\text{x}_\text{i}^2+\sum\text{y}_\text{i}^2)-\Big(\frac{\bar{\text{x}}+\bar{\text{y}}}{2}\Big)^2$
$=\frac{1}{10}(40+105)-9$
$=\frac{55}{10}=\frac{11}{2}$
View full question & answer→MCQ 41 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ If each of the observations $x_1, x_2, ..., X_n $ is increased by a, where ais a negative or positive number, then the variance remains unchanged
Reason $(R):$ Adding or subtracting a positive or negative number to $($or from$)$ each observation of a group does not affect the variance.
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
Assertion Let $x$ be the mean of $x_1, x_2..., x_n.$ Then, variance is given by
$=\sigma_1^2=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}-\bar{\text{x}})^2$
If ais added to each observation, the new observations will be
$\text{y}_\text{i} =\text{x}_\text{i} +\text{a}...(i)$
Let the mean of the new observations be j. Then,
$\bar{\text{y}}=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}\text{y}_\text{i}=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}+\text{a})$
$=\frac{1}{\text{n}}\Bigg[\sum\limits_\text{i=1}^\text{n}\text{x}_\text{i}+\sum\limits_\text{i=1}^\text{n}\text{a}\Bigg]$
$=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}\text{x}_\text{i}+\frac{\text{na}}{\text{n}}=\bar{\text{x}}+\text{a}$
$=\text{i.e.}\ \bar{\text{y}}=\bar{\text{x}}+\text{a}\ ...(ii)$
Thus, the variance of the new observations is
$=\sigma_2^2=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{y}_\text{i}-\bar{\text{y}})^2$
$=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}+\text{a}-\bar{\text{x}}-\text{a})^2$
$[$using Eqs. $(i)$ and $(ii) ]$
$=\frac{1}{\text{n}}\sum\limits_\text{i=1}^\text{n}(\text{x}_\text{i}-\bar{\text{x}})^2=\sigma_1^2$
Thus, the variance of the new observations is same as that of the original observations. Reason We may note that adding $($or subtracting$)$ a positive number to $($or from$)$ each observation of a group does not affect the variance.
View full question & answer→MCQ 51 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Consider the following data
| $x_i$ |
$60$ |
$61$ |
$62$ |
$63$ |
$64$ |
$65$ |
$66$ |
$67$ |
$68$ |
| $f_i$ |
$2$ |
$1$ |
$12$ |
$29$ |
$25$ |
$12$ |
$10$ |
$4$ |
$5$ |
Assertion $(A):$ The mean of the data using shortcut method is $32.$
Reason $(R):$ The standard deviation of the data using shortcut method is $1.69.$ - A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A$.
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
|
Mid value $(x_i)$
|
Frequency $(f_i)$
|
Deviation from mean $d_i= xi - A, A = 64$
|
$d_i^2$
|
$f_id_i $
|
$f_id_i^2$
|
|
$60$
|
$2$
|
$-4$
|
$16$
|
$-8$
|
$32$
|
|
$61$
|
$1$
|
$-3$
|
$9$
|
$-3$
|
$9$
|
|
$62$
|
$12$
|
$-2$
|
$4$
|
$-24$
|
$48$
|
|
$63$
|
$29$
|
$-1$
|
$1$
|
$-29$
|
$29$
|
|
$64$
|
$25$
|
$0$
|
$0$
|
$0$
|
$0$
|
|
$65$
|
$12$
|
$1$
|
$1$
|
$12$
|
$12$
|
|
$66$
|
$10$
|
$2$
|
$4$
|
$20$
|
$40$
|
|
$67$
|
$4$
|
$3$
|
$9$
|
$12$
|
$36$
|
|
$68$
|
$5$
|
$4$
|
$16$
|
$20$
|
$80$
|
|
Total
|
$100$
|
$0$
|
|
$0$
|
$286$
|
Mean $\bar{\text{(x)}}=\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$
$\bar{\text{(x)}}=64\frac{0}{100}=64$
Reason Standard deviatio $(\sigma)$
$=\sqrt{\frac{\sum\text{f}_\text{i}\text{d}_\text{i}^2}{\sum\text{f}_\text{i}}}-\Big(\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2$
$=\sqrt{\frac{286}{100}}-\Big(\frac{0}{100}\Big)^2$
$=\sqrt{286}$
$=1.69$ View full question & answer→MCQ 61 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Consider the following data
| $x_i$ |
$6$ |
$10$ |
$14$ |
$18$ |
$24$ |
$28$ |
$30$ |
| $f_i$ |
$2$ |
$4$ |
$7$ |
$12$ |
$8$ |
$4$ |
$3$ |
Assertion $(A):$ The mean of the data is $19.$
Reason $(R):$ The variance of the data is $43.4.$ - A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- ✓
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: B. $A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
|
$\text{x}_\text{i}$
|
$\text{f}_\text{i}$
|
$\text{x}_\text{i}^2$
|
$\text{f}_\text{i}\text{x}_\text{i}$
|
$\text{f}_\text{i}\text{x}_\text{i}^2$
|
|
$6$
|
$2$
|
$36$
|
$12$
|
$72$
|
|
$10$
|
$4$
|
$100$
|
$40$
|
$400$
|
|
$14$
|
$7$
|
$196$
|
$98$
|
$1372$
|
|
$18$
|
$12$
|
$324$
|
$216$
|
$3888$
|
|
$24$
|
$8$
|
$576$
|
$192$
|
$4608$
|
|
$28$
|
$4$
|
$784$
|
$112$
|
$3136$
|
|
$30$
|
$3$
|
$900$
|
$90$
|
$2700$
|
|
Total $= 130$
|
$40$
|
|
$760$
|
$16176$
|
Mean $\bar{\text{(x})}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{760}{40}=19$
Reason Variance $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}^2}{\sum\text{f}_\text{i}}\Big(\frac{\sum\text{f}_\text{i}\text{x}\text{i}}{\sum\text{f}_\text{i}}\Big)^2$
$=\frac{16176}{40}-\Big(\frac{760}{40}\Big)^2$
$=404.4-(19)^2$
$=404.4-361=43.4$ View full question & answer→MCQ 71 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The weights $($in $kg)$ of $15$ students are as follows $31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30$ If the weight $44\ kg$ is replaced by $46\ kg$ and $27\ kg$ is by $25\ kg,$ then new median is $35.$
Reason $(R)$ The mean deviation from the median of the weights $($in $kg) 54, 50, 40, 42, 51, 45, 47, 57$ is $4.78.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- ✓
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: B. $A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
Assertion Since, $44\ kg$ is replaced by $46\ kg$ and $27\ kg$ is replaced by $25\ kg,$
then the given series becomes $31, 35, 25, 29, 32, 43, 37, 41, 34, 28, 36, 46, 45, 42, 30.$
On arranging this series in ascending order, we get $25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46.$
Total number of students are $15,$ therefore middle term is $8^{th}$ whose corresponding value is $35.$ Reason On arranging the terms in increasing order of magnitude $40, 42, 45, 47, 50, 51, 54, 55, 57.$
Number of terms $N = 9$
$\therefore$ Median $=\Big(\frac{9+1}{2}\Big)$ the term $= 5^{th}$ term $= 50\ kg$
|
Weight $($in $kg)$
|
Deviation median $(d)$
|
$(d)$ |
| $40$ |
$-10$ |
$10$ |
| $42$ |
$-8$ |
$8$ |
| $45$ |
$-5$ |
$5$ |
| $47$ |
$-3$ |
$3$ |
| $50$ |
$0$ |
$0$ |
| $51$ |
$1$ |
$1$ |
| $54$ |
$4$ |
$4$ |
| $55$ |
$5$ |
$5$ |
| $57$ |
$7$ |
$7$ |
|
|
|
$(d) =43$ |
$MD$ from median $=\frac{43}{9}=4.78\text{kg}$ View full question & answer→MCQ 81 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The mean deviation about median calculated for series, where variability is very high, cannot be fully relied.
Reason $(R)$ The median is not a representative of central tendency for the series where degree of variability is very high.
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
Assertion In a series, where the degree of variability is very high, the median is not a representative central tendency.
Thus, the mean deviation about median calculated for such series can not be fully relied.
View full question & answer→MCQ 91 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Consider the following data
| $x_i$ |
$4$ |
$8$ |
$11$ |
$17$ |
$20$ |
$24$ |
$32$ |
| $f_i$ |
$3$ |
$5$ |
$9$ |
$5$ |
$4$ |
$3$ |
$1$ |
Assertion $(A):$ The variance of the data is $45.8.$
Reason $(R):$ The standard deviation of the data is $6.77.$ - A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- ✓
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: B. $A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
|
$\text{x}_\text{i}$
|
$\text{f}_\text{i}$
|
$\text{f}_{\text{i}}\text{x}_\text{i}$
|
$\text{x}_{\text{i}}-\bar{\text{x}}$
|
$(\text{x}_{\text{i}}-\bar{\text{x}})^2$
|
$\text{f}_\text{i}(\text{x}_{\text{i}}-\bar{\text{x}})^2$
|
|
$4$
|
$3$
|
$12$
|
$-10$
|
$100$
|
$300$
|
|
$8$
|
$5$
|
$40$
|
$-6$
|
$36$
|
$180$
|
|
$11$
|
$9$
|
$99$
|
$-3$
|
$9$
|
$81$
|
|
$17$
|
$5$
|
$85$
|
$3$
|
$9$
|
$45$
|
|
$20$
|
$4$
|
$80$
|
$6$
|
$36$
|
$144$
|
|
$24$
|
$3$
|
$72$
|
$10$
|
$100$
|
$300$
|
|
$32$
|
$1$
|
$32$
|
$18$
|
$324$
|
$324$
|
|
|
$30$
|
$420$
|
|
|
$1374$
|
$\text{N}=30\sum\limits_\text{i=1}^7\text{f}_{\text{i}}\text{x}_\text{i}=420$
$=\sum\limits_\text{i=1}^7\text{f}_\text{i}(\text{x}_{\text{i}}-\bar{\text{x}})^2=1374$
$\therefore , \bar{\text{x}}=\frac{\sum\limits_\text{i=1}^7\text{f}_\text{i}\text{x}_{\text{i}}}{\text{N}}=\frac{1}{30}\times420=14$
$\therefore\text{ariance}(\sigma^2)==\sum\limits_\text{i=1}^7\text{f}_\text{i}(\text{x}_{\text{i}}-\bar{\text{x}})^2$
$=\frac{1}{30}\times1374=45.8$
Reason Standard deviation
$=(\sigma)\times\sqrt{45.8}$
$=6.77$ View full question & answer→MCQ 101 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The mean deviation about the mean to find measure of dispersion has certain limitations.
Reason $(R)$ The sum of deviations from the mean is more than the sum of deviations from median. Therefore, the mean deviation about the mean is not very scientific, where degree of variability is very high.
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
Assertion The sum of the deviations from the mean $($minus signs ignored$)$ is more than the sum of the deviations from median. Therefore, the mean deviation about the mean is not very scientific. Thus, in many cases, mean deviation may give unsatisfactory results. Also mean deviation is calculated on the basis of absolute values of the deviations and therefore, cannot be subjected to further algebraic treatment. This implies that we must have some other measure of dispersion. Standard deviation is such a measure of dispersion.
View full question & answer→MCQ 111 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A):$ The proper measure of dispersion about the mean of a set of observations i.e. standard deviation is expressed as positive square root of the variance.
Reason $(R):$ The units of individual observations $x_i$ and the unit of their mean are different that of variance. Since, variance involves sum of squares of $\text{(x}_\text{i} - \bar{\text{x}}).$
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A$.
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A$.
Assertion In the calculation of variance, we find that the units of individual observations $x_i$ and the unit of their mean $\bar{\text{x}}$ are different from that of variance, since variance involves the sum of squares of $\text{(x}_\text{i} - \bar{\text{x}}).$ For this reason, the proper measure of dispersion about the mean of a set of observations is expressed as positive square$-$root of the variance and is called standard deviation. Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.
View full question & answer→MCQ 121 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
If for a distribution $\sum(\text{x-5})=3, \sum(\text{x-5})^2=3$ and the total number ofitems is $18$
Assertion $(A)$ Mean of the distribution is $4.1666.$
Reason $(R)$ Standard deviation of the distribution is $1.54.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
Assertion Given, $\sum\text{(x-5})=3$
$\therefore\sum\text{x}-\sum5=3$
$\Rightarrow\sum\text{x}-5\times18=3\ [\because\text{n}=18]$
$\Rightarrow\sum\text{x}=3+90\Rightarrow\sum\text{x}=93$
Now, $\sum(\text{x}-5)^2=43$
$\Rightarrow\sum(\text{x}^2+25-10\text{x})=43$
$\Rightarrow\sum\text{x}^2+\sum25-10\sum\text{x}=43$
$\Rightarrow\sum\text{x}^2+25\times18-10\times93=43$
$\Rightarrow\sum\text{x}^2=43+930-450$
$\Rightarrow\sum\text{x}^2=973-450$
$\Rightarrow\sum\text{x}^2=523$
Now, mean $=\frac{\sum\text{x}}{\text{n}}=\frac{93}{18}=5.16$
Reason $SD (\sigma)=\sqrt{\frac{\sum\text{x}^2}{\text{n}}-\Big(\frac{\sum\text{x}}{\text{n}}}\Big)^2$
$=\sqrt{\frac{523}{18}-\Big(\frac{93}{18}\Big)}^2$
$=\sqrt{\frac{523\times18-93\times93}{18\times18}}$
$=\frac{1}{18}\sqrt{9414-8649}$
$=\frac{1}{18}\sqrt{765}$
$=\frac{27.66}{18}$
$=1.54$
View full question & answer→MCQ 131 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ In order to find the dispersion of values of $x$ from mean $x$, we take absolute measure of dispersion.
Reason $(R)$ Sum of the deviations from mean $(x)$ is zero.
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
Assertion The deviation of an observation $x$ from a fixed value $'a'$ is the difference $(x - a).$ In order to find the dispersion of values of $x$ from a central value a, we find the deviations about a. An absolute measure of dispersion is the mean of these deviations.
Reason To find the mean, we must obtain the sum of the deviations.
But, we know that a measure of central tendency lies between the maximum and the minimum values of the set of observations.
Therefore, some of the deviations will be negative and some positive.
Thus, the sum of deviations may vanish.
Moreover, the sum of the deviations from mean $\bar{\text{(x)}}$ is zero.
Also, Mean of deviations $=\frac{\text{Sum of deviations }}{\text{Number of observations }}\frac{0}{\text{n}}=0$
Thus, finding the mean of deviations about mean is not of any use for us, as far as the measure of dispersion is concerned.
View full question & answer→MCQ 141 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The variance of first $n$ even natural numbers is $\frac{\text{n}^2-1}{4}.$
Reason $(R)$ The sum of first $2$ natural numbers is $\frac{\text{n}(\text{n+1)}}{2}$ numbers is squares of first n natural numbers is $\frac{\text{n}(\text{n+1)}(2\text{n+1)}}{6}.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
Assertion Sum of $n$ even natural numbers $= n(n + 1)$
mean $\bar{\text{(x)}}=\frac{\text{n(n+1)}}{\text{n}}=\text{n}+1$
Variance $=\Big[\frac{1}{\text{n}}\sum(\text{x}_\text{i})^2\Big]-\bar{\text{(x)}}^2$
$=\frac{1}{\text{n}}\big[2^2+4^2+...+(2\text{n})^2\big]-(\text{n}+1)^2$
$=\frac{1}{\text{n}}\big[2^2+1^2+...+\text{n}^2\big]-(\text{n}+1)^2$
$=\frac{4}{\text{n}}\frac{\text{n(n+1}(2\text{n+1)}}{6}-(\text{n}+1)^2$
$=\frac{(\text{n+1)}[2(2\text{n+1)}-3(\text{n+1)]}}{3}$
$=\frac{(\text{n+1)}(\text{n}-1)}{3}$
$=\frac{\text{n}^2-1}{3}$
View full question & answer→MCQ 151 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)$ The mean deviation about the mean for the data $4, 7, 8, 9, 10, 12, 13, 17$ is $3.$
Reason $(R)$ The mean deviation about the mean for the data $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$ is $8.5.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
$\bar{\text{x}}=\frac{\text{ Sum of terms}}{\text{Number of terms }}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$=\frac{4+7+8+9+10++12+13+17}{8}=10$
|
$\text{x}_\text{i}$
|
$|\text{x}_\text{i}-\bar{\text{x}}|$
|
| $4$ |
$|4 - 10 | = 6$ |
| $7$ |
$|7 -10 | = 3$ |
| $8$ |
$|8 - 10 | = 2$ |
| $9$ |
$|9 - 10 | = 1$ |
| $10$ |
$|10 - 10 | = 0$ |
| $12$ |
$|12 - 10 | = 2$ |
| $13$ |
$|13 - 10 | = 3$ |
| $17$ |
$| 17 - 10 | = 7$ |
|
$\sum\text{x}_\text{i}=80$
|
$\sum|\text{x}_\text{i}-\bar{\text{x}}|=24$
|
$\therefore$ Mean deviation about mean $=\frac{\sum|\text{x}_\text{i}-\bar{\text{x}}|}{\text{n}}=\frac{24}{8}=3$
Reason Mean of the given series
$\bar{\text{x}}=\frac{\text{ Sum of terms}}{\text{Number of terms }}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10}=50$
|
$\text{x}_\text{i}$
|
$|\text{x}_\text{i}-\bar{\text{x}}|$
|
| $38$ |
$| 38 - 50 | = 12$ |
| $70$ |
$| 70 - 50 | = 20$ |
| $48$ |
$| 48 - 50 | = 02$ |
| $40$ |
$| 40 - 50 | = 10$ |
| $42$ |
$| 42 - 50 | = 08$ |
| $55$ |
$| 55 - 50 | =05$ |
| $63$ |
$| 63 - 50 | = 13$ |
| $46$ |
$| 46 - 50 | = 04$ |
| $54$ |
$| 54 - 50 | = 04$ |
| $44$ |
$| 44 - 50 | = 06$ |
|
$\sum\text{x}_\text{i}=500$
|
$\sum|\text{x}_\text{i}-\bar{\text{x}}|=84$
|
$\therefore $ Mean deviation about mean
$=\frac{\sum|\text{x}_\text{i}-\bar{\text{x}|}}{\text{n}}$
$=\frac{84}{10}$
$=8.4$ View full question & answer→