MCQ 11 Mark
Write the correct answer in the following: In a medical examination of students of a class, the following blood groups are recorded:
|
Blood group
|
$A$
|
$AB$
|
$B$
|
$O$
|
|
Number of Student
|
$10$
|
$13$
|
$12$
|
$5$
|
A student is selected at random from the class. The probability that he/ she has blood group $B$, is: - A
$\frac{1}{4}$
- B
$\frac{13}{40}$
- ✓
$\frac{3}{10}$
- D
$\frac{1}{8}$
AnswerCorrect option: C. $\frac{3}{10}$
Total number of students $= 10 + 13 + 12 + 5 = 40$
Let us call event of selected students at random who have blood group $B$ be $E$.
There are $12$ students who has blood group $B$.
So, $\text{P(E)}=\frac{12}{40}=\frac{3}{10}$
View full question & answer→MCQ 21 Mark
Write the correct answer in the following: If $\bar{\text{x}}$ represents the mean of n observations $x_1, x_2, \ldots x_n,$ then value of $\sum\limits_{\text{i}=1}^\text {b} \text{x}_\text{i}-\bar{\text{x}}$ is:
AnswerWe know that algebraic sun of deviations from mean is zero.
$\sum\limits_{\text{a}=1}^\text{b} (\text{x}_\text{t}-\bar{\text{x}})=(\text{x}_1-\bar{\text{x}})+(\text{x}_2-\bar{\text{x}})+(\text{x}_3-\bar{\text{x}})+\ ...\ +(\text{x}_\text{n}-\bar{\text{x}})$
$= (\text{x}_1+\text{x}_2+\text{x}_3+... +\text{x}_\text{n})- \text{n}\bar{\text{x}}$
$\Rightarrow\sum\limits_{\text{t}-1}^\text{b}\text{x}_\text{i}-\text{n}\bar{\text{x}}=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}=0$ $\bigg[\because\sum\limits_{\text{i}=1}^\text{n} \text{x}_\text{i}=\text{n}\bar{\text{x}}\bigg]$
Hence, $(b)$ is correct answer.
View full question & answer→MCQ 31 Mark
Write the correct answer in the following: A grouped frequency table with class intervals of equal sizes using $250-270$ ($270$ not included in this interval) as one of the class interval is constructed for the following data: $268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406, 292, 354, 278, 210, 240, 330, 316, 406, 215, 258, 236$. The frequency of the class $310-330$ is:
AnswerThe observation corresponding to class $310–330$ ($330$ not included in this interval) are $310, 310, 320, 319, 318, 316,$ i.e., $6$ observations.
View full question & answer→MCQ 41 Mark
Write the correct answer in the following: The range of the data: $25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20$ is:
AnswerMaximum value of the variate $= 32$
And the minimum value of the variate $= 6$
Range = Maximum value of the variate-Minimum value of the variate $= 32 - 6 = 26$
View full question & answer→MCQ 51 Mark
Write the correct answer in the following: To draw a histogram to represent the following frequency distribution:
|
Class intervai
|
$5-10$
|
$10-15$
|
$15-25$
|
$25-45$
|
$45-75$
|
|
Frequency
|
$6$
|
$12$
|
$10$
|
$8$
|
$15$
|
The adjusted frequency for the class $25-45$ is: AnswerThe adjusted frequency for the class $25 - 45$ is
$= \frac{\text{Frequency of the class }}{\text{Class width }}\times \text{Minimum width }=\frac{8}{20} \times5=2$
View full question & answer→MCQ 61 Mark
Write the correct answer in the following: For drawing a frequency polygon of a continous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abcissae are respectively:
- A
Upper limits of the classes.
- B
Lower limits of the classes.
- ✓
Class marks of the classes.
- D
Upper limits of perceeding classes.
AnswerCorrect option: C. Class marks of the classes.
Abcissac are the class marks of the classes.
View full question & answer→MCQ 71 Mark
Write the correct answer in the following: The probability that bulbs selected randomly from the lot has life less than $900$ hours is:
- A
$\frac{11}{40}$
- B
$\frac{5}{16}$
- C
$\frac{7}{16}$
- ✓
$\frac{9}{16}$
AnswerCorrect option: D. $\frac{9}{16}$
Total number of bulbs in a lot, $n(S) = 80$
Number of bulbs whose life time is less than $900h, n(E) = 10 + 12 + 23 = 45$
Probability that bulbs has life time less than $900h$ $=\frac{\text{n(E)}}{\text{n(S)}}=\frac{45}{80}=\frac{9}{16}$
Hence, the probability that bulb has life time less than $900$ is $\frac{9}{16}$
View full question & answer→MCQ 81 Mark
Write the correct answer in the following: The width of five continuous classes in a frequency distribution is $5$ and the lower class-limit of the lowest class is $10$. The upper class-limit of the highest class is:
AnswerSol. Width of each of the five continuous classes in a frequency distribution is $5$.
Lower class limit of the lowest class $= 10$
Upper class limit of the lowest class is $10 + 5 = 15$
So, the five continuous classes are
$10 - 15, 15 - 20, 20 - 25, 25 - 30, 30 - 35$
Hence, the upper-class limit of the height class is $35$.
View full question & answer→MCQ 91 Mark
Write the correct answer in the following:In the class intervals $10–20, 20–30,$ the number $20$ is included in:
AnswerCorrect option: B. $20–30$
The number $20$ is included in $20–30$.
Hence, $(b)$ is the correct answer.
View full question & answer→MCQ 101 Mark
Write the correct answer in the following:
A grouped frequency distribution table with classes of equal sizes using $63-72$ ($72$ included) as one of the class is constructed for the following data:
$30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, 40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96, 102, 110, 88, 74, 112, 14, 34, 44$. The number of classes in the distribution will be:
AnswerMinimum value $= 14$
Maximum value $= 112$
The classes are
$13 - 22, 23 - 32, 33 - 42, 43 - 52, 53 - 62, 63 - 72, 73 - 82, 83 - 92, 93 - 102$ and $103 - 112.$
The number of classes in the distribution will be $10$.
View full question & answer→MCQ 111 Mark
Let $\bar{x}$ be thae mean of $x_1, x_2, \ldots ., x_n$, and $\bar{y}$ the mean of $y_1, y_2, \ldots, y_n$. If $\bar{z}$ is the mean of $x_1, x_2$, $\ldots, x_n, y_1, y_2, \ldots, y n$, then $\bar{z}$ is equal to:
- A
$\bar{\text{x}}+\bar{\text{y}}$
- ✓
$\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$
- C
$\frac{\bar{\text{x}}+\bar{\text{y}}}{\text{n}}$
- D
$\frac{\bar{\text{x}}+\bar{\text{y}}}{2\text{n}}$
AnswerCorrect option: B. $\frac{\bar{\text{x}}+\bar{\text{y}}}{2}$
We have $\bar{x}$ is the mean of $x_1, x_2, \ldots \ldots, x_n$, and $\bar{y}$ is the mean of $y_1, y_2, \ldots, y_n$
So, $\overline{\mathrm{x}}=\frac{\left(\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots .+\mathrm{x}_{\mathrm{n}}\right)}{\mathrm{n}}$
$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots \ldots+\mathrm{x}_{\mathrm{n}}=\mathrm{n} \overline{\mathrm{x}}$
And $\overline{\mathrm{y}}=\frac{\left(\mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3+\ldots .+\mathrm{y}_{\mathrm{n}}\right)}{\mathrm{n}}$
$\Rightarrow \mathrm{y}_1+\mathrm{y}_2+\mathrm{y}_3+\ldots+\mathrm{y}_{\mathrm{n}}=\mathrm{n} \overline{\mathrm{y}}$
If $\bar{z}$ is the mean of $x_1, x_2, \ldots, y_1, y_2, \ldots, y_n$ then,
$\overline{\mathrm{z}}=\frac{\mathrm{n} \overline{\mathrm{x}}+\mathrm{n} \overline{\mathrm{y}}}{\mathrm{n}+\mathrm{n}}=\frac{\mathrm{n}(\overline{\mathrm{x}}+\bar{y})}{2 \mathrm{n}}=\frac{\overline{\mathrm{x}}+\overline{\mathrm{y}}}{2}$
View full question & answer→MCQ 121 Mark
Write the correct answer in the following: The mean of $25$ observations is $36$. Out of these observations if the mean of first $13$ observations is $32$ and that of the last $13$ observations is $40$, the $13^{th}$ observation is:
AnswerMean of first $13$ observation $= 32$
$\therefore$ Sum of all first $13$ observation $= (32 \times 13) = 416$
Mean of last $13$ observation $= 40$
$\therefore$ Sum of all last $13$ observation $= (40 \times 13) = 520$
Mean of $25$ observation $= 36$
$\therefore$ Sum of all first $25$ observation $= (36 \times 25) = 900$
Hence, $13^{th}$ observation $= 416 + 520 - 900 = 36$
View full question & answer→MCQ 131 Mark
Write the correct answer in the following: In a sample study of $642$ people, it was found that $514$ people have a high school certificate. If a person is selected at random, the probability that the person has a high school certificate is:
AnswerWe know that empricrical probability $P(E)$ of an event $E$ happen, is given by
$\text{P(E)}=\frac{\text{Number of trials in which the event happed}}{\text{Total number of trials}}$
Let E be the event that a person selected at random has a high school certificate.
$\text{P(E)}=\frac{\text{Numbers of person having high school certificate}}{\text{Total number of people in a sample study}}$
$=\frac{514}{642}=0.8$
View full question & answer→MCQ 141 Mark
Write the correct answer in the following: The median of the data $78, 56, 22, 34, 45, 54, 39, 68, 54, 84$ is:
AnswerArranging the data in ascending order, we get
$22, 34, 39, 45, 54, 56, 78, 84$
Here $n = 9$, which is an odd number.
$\therefore\ \text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}$
$\text{Value}=\Big(\frac{9+1}{2}\Big)^{\text{th}}$
$\text{value}=5^{\text{th}}\text{value}$
So, median $= 54$
View full question & answer→MCQ 151 Mark
Write the correct answer in the following:
The class marks of a frequency distribution are given as follows: $15, 20, 25, ....$ The class corresponding to the class mark $20$ is:
- A
$12.5 - 17.5$
- ✓
$17.5 - 22.5$
- C
$18.5 - 21.5$
- D
$19.5 - 20.5$
AnswerCorrect option: B. $17.5 - 22.5$
The class mark are $15, 20, 25, ….$
The size of each class interval is $25 - 20 = 20 - 15 = 5$
Hence, the class interval corresponding to the class mark $20$ is,
$(20 - 2.5) - (20 + 2.5)$ i.e., $17.5 - 22.5.$
So, $(b)$ is the correct answer.
View full question & answer→MCQ 161 Mark
Write the correct answer in the following: Median of the following numbers, $4, 4, 5, 7, 6, 7, 7, 12, 3$ is:
AnswerFirst, we arrange the given numbers in ascending order is,
$3, 4, 4, 5, 6, 7, 7, 7$ and $12$
Here, $n = 9$
Since, $n$ is odd, so we use the formula for median,
Now, Median $=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{observation}$
$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{observation}$ [Put $n = 9$]
$=\Big(\frac{10}{2}\Big)^{\text{th}}\text{observation}$
$=5^{\text{th}}\text{observation}=6$
View full question & answer→MCQ 171 Mark
Write the correct answer in the following:
$80$ bulbs are selected at random from a lot and their life time (in hrs) is recorded in the form of a frequency table given below:
| Life time(in hours) |
$300$
|
$500$
|
$700$
|
$900$ |
$1100$ |
|
Frequency
|
$10$ |
$12$
|
$23$
|
$25$ |
$10$ |
One bulb is selected at random from the lot. The probability that its life is $1150$ hours, is: - A
$\frac{1}{80}$
- B
$\frac{7}{16}$
- ✓
$0$
- D
$1$
AnswerTotal number of bulbs $= 80$
Let $E$ of the event that bulb selected at random form the lot has life time $1150\ hrs$.
We see from the frequency table given above that none of the bulb has life time $1150\ hrs$.
$\therefore\ \text{P(E)}=\frac{0}{80}=0$
View full question & answer→MCQ 181 Mark
Write the correct answer in the following:
There are $50$ numbers. Each number is subtracted from $53$ and the mean of the numbers so obtained is found to be $-3.5$. The mean of the given numbers is:
- A
$46.5$
- B
$49.5$
- C
$53.5$
- ✓
$56.5$
AnswerCorrect option: D. $56.5$
Given, $n = 50$, then mean $\bar{\text{x}}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$
Then, $\bar{\text{x}}=\frac{1}{50}\times\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}$
$\Rightarrow\sum\limits^{50}_{\text{i}=1}\text{x}_\text{i}=50\bar{\text{x}}$
Now, subtract each observation from $53$, we get a new mean say $\bar{\text{x}}_\text{new}$
$\therefore\ \bar{\text{x}}_\text{new}=\frac{(-\text{x}_1+53)+(-\text{x}_2+53)\ +\ .....\ +(-\text{x}_{50}+53)}{50}$
$\Rightarrow-3.5=\frac{-(\text{x}_1+\text{x}_2\ +\ .....\ +\text{x}_{50})+(53+53\ +\ ....\ +50\text{times})}{50}$
$\Rightarrow-3.5\times50=(-\text{x}_1+\text{x}_2+\ ....\ +\text{x}_{50})+53\times50$
$\therefore$ Mean of $50$ observation $=\frac{1}{50}\sum\limits^{50}_{\text{i}=1}\text{x}_{\text{i}}$
$=\frac{1}{50}\times2852=56.5$ $\Bigg[\because\text{ mean}=\frac{\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}\Bigg]$
Hence, the mean of given number is $56.5$.
View full question & answer→MCQ 191 Mark
Write the correct answer in the following: Let $m$ be the mid-point and $l$ be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
- A
$2m + l$
- ✓
$2m - l$
- C
$m - l$
- D
$m - 2l$
AnswerCorrect option: B. $2m - l$
Let $x$ and $y$ be the lower and upper class limit of a continuous frequency distribution.
Now, mid-point of a class $=\frac{(\text{x}+\text{y})}{2}=\text{m}$ [given]
$\Rightarrow x + y = 2m = x + l = 2m$
$\big[\therefore$ $y = l$ = upper class limit (given)$\big]$
$\Rightarrow x = 2m - l$
Hence, the lower class limit of the class is $2m - l$.
View full question & answer→MCQ 201 Mark
Write the correct answer in the following:
If each observation of the data is increased by $5$, then their mean:
AnswerCorrect option: D. Is increased by $5$.
Let $x_1, x_2, \ldots, x_n$ be the $n$ observation,
Then, old mean $\bar{\text{x}}_{\text{old}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}\ \dots(\text{i})$
Now, adding $5$ in each obsevation, the new mean becomes
$\bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+5)+(\text{x}_2+5)+\ ...\ +(\text{x}_\text{n}+5)}{\text{n}}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{(\text{x}_1+\text{x}_2+\ ...\ +\text{x}_\text{n})+5\text{n}}{}$
$\Rightarrow\ \bar{\text{x}}_{\text{new}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{x}_\text{i}}{\text{n}}+5=\bar{\text{x}}_{\text{old}}+5$ [from Eq. $(i)$]
$\Rightarrow\ \bar{\text{x}}_\text{new}=\bar{\text{x}}_{\text{old}}+5$
Hence, the new mean is increased by $5$.
View full question & answer→MCQ 211 Mark
Write the correct answer in the following: In a survey of $364$ children aged $19-36$ months, it was found that $91$ liked to eat potato chips. If a child is selected at random, the probability that he/ she does not like to eat potato chips is:
- A
$0.25$
- B
$0.50$
- ✓
$0.75$
- D
$0.80$
AnswerCorrect option: C. $0.75$
Total number of survey children’s age from $19 - 36$ months, $n(S) = 364$ In those of them $91$ out of them liked to eat potato chips.
$\therefore$ Number of children who do not like to eat potato chips, $n(E) = 364 - 91 = 273$
$\therefore$ Probability that he/she does not like to eat potato chips $=\frac{\text{n(E)}}{\text{n(s)}}=\frac{273}{364}=0.75$
Hence, the probability that he/ she does not like to eat potato chips is $0.75$.
View full question & answer→MCQ 221 Mark
Write the correct answer in the following: If the mean of the observations: $x, x + 3, x + 5, x + 7, x + 10$ is $9$, the mean of the last three observations is:
- A
$10\frac{1}{3}$
- B
$10\frac{2}{3}$
- ✓
$11\frac{1}{3}$
- D
$11\frac{2}{3}$
AnswerCorrect option: C. $11\frac{1}{3}$
Given that, the mean of the observation $x, x + 3, x + 5, x + 7$ and $x + 10$ is $9$.
$\therefore\ \frac{\text{x}+\text{x}+3+\text{x}+5+\text{x}+7+\text{x}+10}{5}=9$
$\Rightarrow {\text{5x}}+25=45$
$\Rightarrow\text{5x}=20$
$\Rightarrow \text{x}=4$
$\therefore$ Last three odservations are $x + 5 = 4 + 5 = 9, x + 7 = 4 + 7 = 11$
and $x + 10 = 4 + 10 = 14$
So, the mean of last three observations $=\frac{9+11+14}{3}=\frac{34}{3}=11\frac{1}{3}$
Hence, the mean of last three observation is $11\frac{1}{3}.$
View full question & answer→MCQ 231 Mark
Write the correct answer in the following: Mode of the data $15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15$ is:
AnswerWe first arrange the given data in ascending order as follows $14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20$ From above, we see that $15$ occurs most frequently i.e., $5$ times.
View full question & answer→MCQ 241 Mark
Write the correct answer in the following:
The class mark of the class $90-120$ is:
Answer$\text{Class mark }=\frac{\text{Upperlimit + Lowerlimit}}{2}$
$\Rightarrow \text{Class mark}=\frac{90+120}{2}=\frac{210}{2}=105$
View full question & answer→MCQ 251 Mark
If $\bar{\text{x}}$ is the mean of $x_1, x_2, \ldots . ., x_n$, then for $\text{a}\ne0,$ then mean of $\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n},\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}$ is:
- A
$\Big(\text{a}+\frac{1}{\text{a}}\Big)\bar{\text{x}}$
- ✓
$\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}$
- C
$\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{\text{n}}$
- D
$\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\bar{\text{x}}}{2\text{n}}$
AnswerCorrect option: B. $\Big(\text{a}+\frac{1}{\text{a}}\Big)\frac{\bar{\text{x}}}{2}$
Given, mean of $x_1, x_2, \ldots . ., x_n$ is $\bar{\text{x}}$
$\therefore\ \sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}=\text{n}\bar{\text{x}}$
Now, let the mean of $\Big(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n},\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)$ is $\bar{\text{z}}$
Then, $\bar{\text{z}}=\frac{(\text{ax}_1,\text{ax}_2,\ ....\text{ ax}_\text{n})+\Big(\frac{\text{x}_1}{\text{a}},\frac{\text{x}_2}{\text{a}},\ ...., \ \frac{\text{x}_\text{n}}{\text{a}}\Big)}{\text{n}+\text{n}}$
$=\frac{\text{a}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})+\frac{1}{\text{a}}(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)(\text{x}_1+\text{x}_2+\ ....\ +\text{x}_\text{n})}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big)\cdot\text{n}\bar{\text{x}}}{2\text{n}}$
$=\frac{\big(\text{a}+\frac{1}{\text{a}}\big){\bar{\text{x}}}}{2}$
View full question & answer→MCQ 261 Mark
The mean of $100$ observations is $50$. If one of the observations which was $50$ is replaced by $150$, the resulting mean will be:
AnswerWe have, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow\sum\text{x}_\text{i}=50\times100=5,000$
If one of the observation which was $50$ is resplaced by $150$, then
$\sum\text{x}_\text{i}=5,000-50+150=5100$
Then, the resulting mean $=\frac{5100}{100}=51$
View full question & answer→MCQ 271 Mark
Write the correct answer in the following: The mean of five numbers is $30$. If one number is excluded, their mean becomes $28$. The excluded number is:
AnswerLet $x_1, x_2, x_3, x_4$, and $x_5$ be five numbers and one of the excluded number be $x_5$,
Given, mean of the numbers $= 30$
$\Rightarrow\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}{5}=30$
$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5}=150$
$\Rightarrow {\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}=150-\text{x}_5$
On dividing both sides by $4$, we get
$\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4}{4}=\frac{150-\text{x}_5}{4}\ \dots(\text{i})$
Given, mean of the numbers $= 28$
$\therefore\ \frac{150-\text{x}_5}{4}=28$ [from Eq. $(i)$]
$\Rightarrow 150 -\text{x}_5=112$
$\Rightarrow \text{x}_5=150 -112$
$\Rightarrow\text{x}_5=38$
Hence, the excluded number is $38$.
View full question & answer→MCQ 281 Mark
Write the correct answer in the following:
In a frequency distribution, the mid value of a class is $10$ and the width of the class is. The lower limit of the class is:
AnswerLet x and y be the uppel and lower and lower class limit in a frequency distribution.
Now, mid value of a class $\frac{(\text{x}+\text{y})}{2}=10$ [given]
$\Rightarrow x + y = 20 ...(i)$
Also, given that, width of class $x - y = 6 ...(ii)$
On adding Eqs. $(i)$ and $(ii)$, we get
$2x = 20 + 6$
$\Rightarrow 2x = 26$
$\Rightarrow x = 13$
On putting $x = 13$ in Eq. $(i)$, we get
$13 + y = 20$
$\Rightarrow y = 7$
Hence, the lower limit of the class is $7$.
View full question & answer→MCQ 291 Mark
If $\bar{\text{x}}_1, \bar{\text{x}}_2, \bar{\text{x}}_3, ....., \bar{\text{x}}_\text{n}$ are the means of n group with $\mathrm{n}_1, \mathrm{n}_2, \ldots, \mathrm{n}_{\mathrm{n}}$ number of observations respectively, then the mean $\bar{\text{x}}$ of all the groups taken together is given by:
- A
$\sum\limits^\text{n}_{\text{i}=1}$
- B
$\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}^2}$
- ✓
$\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}_2}$
- D
$\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{2\text{n}}$
AnswerCorrect option: C. $\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\text{n}_2}$
If $\bar{\text{x}}_1, \bar{\text{x}}_2, \bar{\text{x}}_3, ....., \bar{\text{x}}_\text{n}$ are the mean of n group with $\mathrm{n}_1, \mathrm{n}_2, \ldots, \mathrm{n}_{\mathrm{n}}$ number of observation respectively, then the mean $\bar{\text{x}}$
$\bar{\text{x}}=\frac{\text{n}_1\bar{\text{x}}_1+\text{n}_2\bar{\text{x}}_2+\text{n}_3\bar{\text{x}}_3+\ ....\ +\text{n}_\text{n}\bar{\text{x}}_\text{n}}{\text{n}_1+\text{n}_2+\text{n}_3+\ ....\ +\text{n}_\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}}$
Hence, the mean $\bar{\text{x}}$ of all group taken together is gien up $=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}\bar{\text{x}}_\text{i}}{\sum\limits^\text{n}_{\text{i}=1}\text{n}_\text{i}}$
View full question & answer→MCQ 301 Mark
Write the correct answer in the following: Two coins are tossed $1000$ times and the outcomes are recorded as below:
|
Number of heads
|
$2$
|
$1$
|
$0$
|
|
Frequency
|
$200$
|
$550$
|
$250$
|
Based on this information, the probability for at most one head is: - A
$\frac{1}{5}$
- B
$\frac{1}{4}$
- ✓
$\frac{4}{5}$
- D
$\frac{3}{4}$
AnswerCorrect option: C. $\frac{4}{5}$
The total number of coins tossed, $n(S) = 1000$
Number of outcomes in which atmost one head, $n(E) = 550 + 250 = 800$
$=\frac{\text{n(E)}}{\text{n(S)}}=\frac{800}{1000}=\frac{4}{5}$
Hence, the probability for atmost one head is $\frac{4}{5}$
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