MCQ 1011 Mark
The mean of $n$ observation is $\bar{\text{X}}$. If the first item is increased by $1,$ second by $2$ and so on, then the new mean is :
AnswerCorrect option: C. $\overline{\text{X}}+\frac{\text{n}+1}{2}$
Let $x_1, x_2, x_3, \ldots . . . ., x_n$ be the $n$ observations.
$\text{Mean}=\overline{\text{X}}=\frac{\text{x}_1+\text{x}_2+......+\text{x}_\text{n}}{2}$
$\Rightarrow\text{x}_1+\text{x}_2+\text{x}_3+......\text{x}_\text{n}=\text{n}\overline{\text{X}}$
if the first item is increased by $1,$ second by $2$ and so on.
Then, the new observations are $x_1+1, x_2+2, x_3+3, \ldots . x_n+n$
$\text{New mean}=\frac{(\text{x}_1+1)+(\text{x}_2+2)+(\text{x}_3+3)+.....+(\text{x}_\text{n}+\text{n})}{\text{n}}$
$=\frac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_\text{n}+(1+2+3+...+\text{n})}{\text{n}}$
$=\frac{\text{n}\overline{\text{X}}+\frac{\text{n}(\text{n+1})}{2}}{\text{n}}$
$\overline{\text{X}}+\frac{\text{n}+1}{2}$
View full question & answer→MCQ 1021 Mark
The arithmetic mean of $\{1, 2, 3, ..., n\}$ is :
- ✓
$\frac{\text{n}+1}{2}$
- B
$\frac{\text{n}-1}{2}$
- C
$\frac{\text{n}}{2}$
- D
$\frac{\text{n}}{2}+1$
AnswerCorrect option: A. $\frac{\text{n}+1}{2}$
Arithmetic mean of $\{1, 2, 3, ......, n\}$
$=\frac{1+2+3+......+\text{n}}{\text{n}}$
$=\frac{\frac{\text{n}(\text{n}+1)}{2}}{\text{n}}$
$=\frac{\text{n}+1}{2}$
Hence, the correct option is $(a)$
View full question & answer→MCQ 1031 Mark
If the arithmetic mean of $\{7, 8, x, 11, 14\}$ is $x,$ then $x =$
AnswerArithmetic mean of $\{7, 8, x, 11, 14\},$ is $x$
$\Rightarrow\frac{7+8+\text{x}+11+14}{5}=\text{x}$
$\Rightarrow\frac{40+\text{x}}{5}=\text{x}$
$\Rightarrow40+\text{x}=5\text{x}$
$\Rightarrow5\text{x}-\text{x}=40$
$\Rightarrow4\text{x}=40$
$\Rightarrow\text{x}=\frac{40}{4}=10$
View full question & answer→MCQ 1041 Mark
In the formul $\bar{\text{x}}=\text{a+h}\big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\big)$ a stands for :
AnswerIn the formula $\bar{\text{x}}=\text{a+h}\big(\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\big)$
$a\ ’$ stands for Assumed Mean
i.e., assumed mean value is the mid value $(\text{x}_\text{i})$ of class intervals of a set of grouped data.
View full question & answer→MCQ 1051 Mark
Which one of the following measures is determined only after the construction of cumulative frequency distribution?
AnswerThe cumulative frequency table is useful in determining the median.
View full question & answer→MCQ 1061 Mark
Choose the correct answer from the given four options : While computing mean of grouped data, we assume that the frequencies are :
- A
Evenly distributed over all the classes.
- B
Centred at the classmarks of the classes.
- ✓
Centred at the upper limits of the classes.
- D
Centred at the lower limits of the classes.
AnswerCorrect option: C. Centred at the upper limits of the classes.
In computing the mean of grouped data, the frequencies are centred at the class marks of the classes.
View full question & answer→MCQ 1071 Mark
Consider the data :
|
Classes
|
$65-85$ |
$85-105$ |
$105-125$ |
$125-145$ |
$145-165$ |
$165-185$ |
$185-205$ |
|
Frequency
|
$4$ |
$5$ |
$18$ |
$20$ |
$17$ |
$7$ |
$4$ |
The difference of the upper limit of the median class and the lower limit of the modal class is : Answer
|
Classes
|
$65-85$ |
$85-105$ |
$105-125$ |
$125-145$ |
$145-165$ |
$165-185$ |
$185-205$ |
|
Frequency
|
$4$ |
$5$ |
$18$ |
$20$ |
$17$ |
$7$ |
$4$ |
|
Cumulative Frequency
|
$4$ |
$9$ |
$27$ |
$47$ |
$64$ |
$71$ |
$75$ |
Here $N = 75$
$\Rightarrow\frac{\text{N}}{2}=37.5$
$\therefore$ The median class is $125-145$ and Modal class is $125-145$
The difference of upper limits of Median class and lower limit of Modal class $= 145 - 125 = 20$ View full question & answer→MCQ 1081 Mark
Construction of cumulative frequency table is useful to determine :
AnswerA cumulative frequency distribution is the sum of the class and all classes below it in a frequency distribution. Construction of cumulative frequency table is useful to determine Median.
View full question & answer→MCQ 1091 Mark
If $\sum\text{f}_\text{i}\text{s}_\text{i}=1860$ and $ \sum\text{f}_\text{i}=30$ then the value of $\bar{\text{x}}$ is :
AnswerIn the formula $\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}},\text{d}_\text{i}$ represents.
$\bar{\text{x}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{1860}{30}=62$
View full question & answer→MCQ 1101 Mark
Which of the following is not a measure of central tendency ?
AnswerThe most common measures of central tendency are mean, median and mode.
Standard deviation is a measure of the dispersion of a set of data from its mean.
It is calculated as the square root of variance.
Hence standard deviation is not a measure of central tendency.
View full question & answer→MCQ 1111 Mark
The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data gives its :
AnswerThe less than ogive and more than ogive when drawn on the same graph intersect at a point.
From this point, if we draw a perpendicular on the $x\ -$ axis, the point at which it cuts the $x\ -$ axis gives us the median.
Thus, the abscissa of the point of intersection of less than type and of the more than type cumulative curves of a grouped data gives its median.
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 1121 Mark
Mode of $\{2, 3, 4, 5, 0, 1, 3, 3, 4, 3\}$ is $3$ :
AnswerMode is the term which appears maximum number of times.
The terms are : $\{2, 3, 4, 5, 0, 1, 3, 3, 4, 3\}.$
Arranging them in ascending order : $\{0, 1, 2, 3, 3, 3, 3, 4, 4, 5\}.$
$3$ is occurring maximum number of times. hence mode is $3.$
View full question & answer→MCQ 1131 Mark
If the median of the data : $\{ \text{6, 7, x – 2, x, 17, 20},\}$ written in ascending order, is $16$. Then $x =$
AnswerMedian of $\{6, 7, x - 2, x, 17, 20\}$ is $16$
Here $n = 6$
$\therefore\text{Median}=\frac{1}{2}\Big[\frac{\text{n}}{2}\text{th}+\Big(\frac{\text{n}}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}\Big[\frac{6}{2}+\Big(\frac{6}{2}+1\Big)\Big]\text{term}$
$=\frac{1}{2}(3\text{rd}+4\text{th})\text{term}$
$=\frac{1}{2}(\text{x}-2+\text{x})$
$=\frac{1}{2}(2\text{x}-2)=\text{x}-1$
$\therefore\text{x}-1=16$
$\Rightarrow\text{x}=16+1=17$
View full question & answer→MCQ 1141 Mark
$\sum(\text{x}_\text{i}-\bar{\text{x}})$ is equal to :
Answer$\sum(\text{x}_\text{i}-\bar{\text{x}})=\sum(\text{x}_\text{i}-\sum(\bar{\text{x}})$
$\sum(\bar{\text{x}})=\sum(\bar{\text{n}})$ by defination
But $(\bar{\text{x}})=\frac{\sum{\text{x}_\text{i}}}{\text{n}}$ by defination
$\text{so},=\sum\text{x}_\text{i}-\sum\text{x}_\text{i}-\text{n}\frac{(\sum\text{s}_{\text{i}})}{\text{n}}$
which is equal to $=(\sum\text{x}_\text{i})-(\sum\text{x}_\text{i})=0$
View full question & answer→MCQ 1151 Mark
While computing mean of grouped data, we assume that the frequencies are :
- A
Evenly distributed over all the classes.
- ✓
Centred at the class marks of the classes.
- C
Centred at the upper limit of the classes.
- D
Centred at the lower limit of the classes.
AnswerCorrect option: B. Centred at the class marks of the classes.
We know that while computing the mean of a grouped data,
the frequencies are centered at the class marks of the classes.
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 1161 Mark
Mean of a certain number of observation is. If each observation is divided by $m(m \neq 0)$ and increased by $n,$ then the mean of new observation is :
- ✓
$\frac{\bar{\text{x}}}{\text{m}}+\text{n}$
- B
$\frac{\bar{\text{x}}}{\text{n}}+\text{m}$
- C
$\bar{\text{x}}+\frac{\text{n}}{\text{m}}$
- D
$\bar{\text{x}}+\frac{\text{m}}{\text{n}}$
AnswerCorrect option: A. $\frac{\bar{\text{x}}}{\text{m}}+\text{n}$
Let $\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3, \ldots \ldots, \mathrm{y}_{\mathrm{k}}$ be $k$ observations.
Mean of the observation $=\bar{\text{x}}$.
$\Rightarrow\frac{\text{y}_1+\text{y}_2+\text{y}_3+...\text{y}_\text{k}}{\text{k}}=\bar{\text{x}}$
$\Rightarrow\text{y}_1+\text{y}_2+\text{y}_3+.....\text{y}_\text{k}=\text{k}\bar{\text{x}}\ .......(1)$
each observation is divided by $m$ and increased by $n,$ then the new observations are
$\frac{\text{y}_1}{\text{m}}+\text{n},\frac{\text{y}_2}{\text{m}}+\text{n},\frac{\text{y}_3}{\text{m}}+\text{n},.....,\frac{\text{y}_\text{k}}{\text{m}}+\text{n}$
$\therefore$ Mean of new observations
$=\frac{\Big(\frac{\text{y}_1}{\text{m}}+\text{n}\Big)+\Big(\frac{\text{y}_2}{\text{m}}+\text{n}\Big)+.....+\Big(\frac{\text{y}_\text{k}}{\text{m}}+\text{n}\Big)}{\text{k}}$
$=\frac{\Big(\frac{\text{y}_1}{\text{m}}+\frac{\text{y}_2}{\text{m}}+.....+\frac{\text{y}_\text{k}}{\text{m}}+\Big)+(\text{n}+\text{n}+.....+\text{n})}{\text{k}}$
$=\frac{\text{y}_1+\text{y}_2+....+\text{y}_\text{k}}{\text{mk}}+\frac{\text{nk}}{\text{k}}$
$=\frac{\text{k}\bar{\text{x}}}{\text{mk}}+\frac{\text{nk}}{\text{k}}$
$=\frac{\bar{\text{x}}}{\text{m}}+\text{n}$
View full question & answer→MCQ 1171 Mark
One of the methods of determining mode is :
- A
Mode $= 2$ Median $– \ 3$ Mean.
- B
Mode $= 2$ Median $+ \ 3$ Mean.
- ✓
Mode $= 3$ Median $–\ 2$ Mean.
- D
Mode $= 3 $ Median $+ \ 2$ Mean.
AnswerCorrect option: C. Mode $= 3$ Median $–\ 2$ Mean.
Mode $= 3$ Median $– \ 2$ Mean.
View full question & answer→MCQ 1181 Mark
and the mean of the distribution is $7,$ then the value of $p$ is :
AnswerSince, Mean $=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$
$\Rightarrow7=\frac{4\text{p}+63}{17}$
$\Rightarrow\text{4p}+63=119$
$\Rightarrow\text{4p}=119-63$
$\Rightarrow\text{4p}=56$
$\Rightarrow\text{p}=14$
View full question & answer→MCQ 1191 Mark
To represent the more than type graphically, we plot the $.........$ on the $x-$ axis.
AnswerThe lower limit for every class is the smallest value in that class on the other hand the upper limit for every class is the greatest value in that class.
To represent ‘the more than type’ graphically, we plot the lower limits on the $x\ -$ axis and cumulative frequency on the $y\ -$ axis to find the median.
View full question & answer→MCQ 1201 Mark
Consider the following frequency distribution :
| Class: |
$0-5$ |
$6-11$ |
$12-17$ |
$18-23$ |
$24-29$ |
| Frequency: |
$13$ |
$10$ |
$15$ |
$8$ |
$11$ |
The upper limit of the median class is : AnswerCorrect option: B. $17.5$
Given, classes are not continuous, so we make continuous by subtracting $0.5$ from lower limit and adding $0.5$ to upper limit of each class.
|
Class
|
Frequency
|
Cumulative Frequency
|
| $0.5-5.5$ |
$13$ |
$13$ |
| $5.5-11.5$ |
$10$ |
$23$ |
| $11.5-17.5$ |
$15$ |
$38$ |
| $17.5-23.5$ |
$8$ |
$46$ |
| $23.5-29.5$ |
$11$ |
$57$ |
Here,$\frac{\text{N}}{2}=\frac{57}{2}=28.5,$ which lies in the interval $11.5-17.5.$
Hence, the upper limit is $17.5.$ View full question & answer→MCQ 1211 Mark
The wickets taken by a bowler in $10$ cricket matches are $\{2, 6, 4, 5, 0, 3, 1, 3, 2, 3.\}$ The median of the data is :
AnswerArranging the given data in ascending order,
$\{0, 1, 2, 2, 3, 3, 3, 4, 5, 6\}$
Here $n = 10,$ which is even.
$\therefore$ Median $=\frac{1}{2}\Bigg[\Big(\frac{\text{n}}{2}\Big)^\text{th} \text{term}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{term}\Bigg]$
$=\frac{1}{2}\big[5^\text{th}\text{ term}+6^\text{th}\text{ term}\big]$
$\Rightarrow\frac{1}{2}\big[3+3\big]=\frac{6}{2}$
$=3$
View full question & answer→MCQ 1221 Mark
Choose the correct answer from the given four options : Consider the following frequency distribution :
| Class |
$0-5$ |
$6-11$ |
$12-27$ |
$18-23$ |
$24-29$ |
| Frequency |
$13$ |
$10$ |
$15$ |
$8$ |
$11$ |
The upper limit of the median class is : AnswerCorrect option: B. $17.5$
Here,
| Class |
Frequency |
Cumulative frequency |
| $-0.5-5.5$ |
$13$ |
$13$ |
| $5.5-11.5$ |
$10$ |
$23$ |
| $11.5-17.5$ |
$15$ |
$38$ |
| $17.5-23.5$ |
$8$ |
$46$ |
| $23.5-29.5$ |
$11$ |
$57$ |
View full question & answer→MCQ 1231 Mark
The mode of a frequency distribution is obtained graphically from :
AnswerMode can be obtained graphically from a histogram.
View full question & answer→MCQ 1241 Mark
The mean of all the factors of $24$ is :
AnswerAll factors of $24$ are $\{1, 2, 3, 4, 6, 8, 12, 24\}$
$\therefore\text{Mean}=\frac{\text{Sum of all factors of 24 }}{\text{Number of factours of 24}}$
$=\frac{1+2+3+4+6+8+12+24}{8}$
$=\frac{60}{8}$
$=7.5$
View full question & answer→MCQ 1251 Mark
The mean of the first $10$ natural numbers is $-$
AnswerThe first $10$ natural numbers are $1, 2, 3, ... 10$
$\therefore\text{Mean}=\frac{\text{sum of first 10 natural numbers are 1,2,3....10}}{10}$
$=\frac{1+2+3+.....+10}{10}$
$=\frac{55}{10}$
$=5.5$
View full question & answer→MCQ 1261 Mark
For any collection of $n$ items,$\Big(\sum\text{x}\Big)-\overline{\text{x}}=$
- A
$\text{n}\overline{\text{x}}$
- B
$(\text{n+1)}\overline{\text{x}}$
- ✓
$(\text{n}-1)\overline{\text{x}}$
- D
$0$
AnswerCorrect option: C. $(\text{n}-1)\overline{\text{x}}$
For any collection of $'n\ ’$ items, $\Big(\sum\text{x}\Big)-\overline{\text{x}}=\text{n}{\big(\overline{\text{x}}}\big)-\overline{\text{x}}=\big(\text{n}-1\big)$
View full question & answer→MCQ 1271 Mark
The median and mode of a frequency distribution are $26$ and $29$ respectively. Then, the mean is :
- A
$27.5$
- ✓
$24.5$
- C
$28.4$
- D
$25.8$
AnswerCorrect option: B. $24.5$
Median $= 26$
Mode $= 29$
Mode $= 3$ median $- 2$ mean
Hence, Mean $=\frac{3\text{Median}-\text{Mode}}{2}$
$=\frac{3(26)-29}{2}$
$=\frac{78-29}{2}$
$=\frac{49}{2}$
$=24.5$
View full question & answer→MCQ 1281 Mark
The mean of first $n$ odd natural number is :
- A
$\frac{\text{n}+1}{2}$
- B
$\frac{\text{n}}{2}$
- ✓
$\text{n}$
- D
$\text{n}^2$
AnswerCorrect option: C. $\text{n}$
Mean of first $n$ odd numbers Sum of first $n$ odd number
$=\frac{\text{n}}{2}[2\text{a}+(\text{n}+1)\text{d}]$
i.e. $\{1 + 3 + 5 + 7 + ........ n\}$ term
$=\frac{\text{n}}{2}[2\times1+(\text{n}+1)\times2]$
$($Here $a = 1, d = 2)$
$=\frac{\text{n}}{2}[2+2\text{n}-2]=\frac{\text{n}}{2}\times2\text{n}=\text{n}^2$
$\therefore\text{Mean}=\frac{\text{Sum of n terms}}{\text{n}}=\frac{\text{n}^2}{\text{n}}=\text{n}$
View full question & answer→MCQ 1291 Mark
If the arithmetic mean of $x, x + 3, x + 6, x + 9,$ and $x + 12$ is $10$, then $x =$
AnswerMean of ${x,x}+3,{x}+6,{x}+9,{x}+12=10$
$\Rightarrow\frac{\text{x}+\text{x}+3+\text{x}+6+\text{x}+9+\text{x}+12}{5}=10$
$\Rightarrow\frac{5\text{x}+30}{5}=10$
$\Rightarrow\text{x}+6=10$
$\Rightarrow\text{x}=10-6=4$
View full question & answer→MCQ 1301 Mark
The wickets taken by a bowler in $10$ cricket matches are $\{2, 6, 4, 5, 0, 2, 1, 3, 2, 3\}$. The median of the data is :
AnswerArranging the given data in ascending order, we get
$\{0, 1, 2, 2, 2, 3, 3, 4, 5, 6\}$
Here $, n = 10,$ which is even.
$\therefore$ Median $=\frac{1}{2}\Big[\big(\frac{\text{n}}{2}\big)^\text{th}\text{ term}+\big(\frac{\text{n}}{2}+1\big)^\text{th}\text{ term}\Big]$
$=\frac{1}{2}\ [5^{th}$ term $+ 6^{th}$ term $]$
$\Rightarrow $ Median $=\frac{1}{2}[2+5]=\frac{5}{2}$
$= 2.5$
View full question & answer→MCQ 1311 Mark
In a data, if $~l=60, h=15, f_1=16, f_0=6, f_2=6$, then the mode is :
AnswerCorrect option: A. $67.5$
Mode $l =\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{f}_1-]\text{f}_0-\text{f}_2}\Big)\times\text{h}$
$=60+\frac{16-6}{2\times16-6-6}\times15$
$=60+\frac{10-6}{32-12}\times15$
$=60+\frac{10}{20}\times15$
$=60+7.5$
$=67.5$
View full question & answer→MCQ 1321 Mark
if $\text{u}_\text{i}=\frac{\text{x}_\text{i}-25}{10}\sum\text{f}_\text{i}\text{u}_\text{i}=20,\sum\text{f}_\text{i}=100,$ then $\bar{\text{x}}$
Answer$\text{u}_\text{i}=\frac{\text{x}_\text{i}-25}{10},\ \sum\text{f}_\text{i}\text{u}_\text{i}=20,\ \sum\text{f}_\text{i}=100$
Here assumed mean $= 25$
and class interval $(h) = 10$
$\therefore\ \bar{\text{x}}=\text{A}+\frac{\sum\text{f}_\text{i}\text{u}_\text{i}}{\sum\text{f}_\text{i}}\times\text{h}$
$=25+\frac{20}{100}\times10$
$=25+2=27$
View full question & answer→MCQ 1331 Mark
The marks obtained by $9$ students in Mathematics are $\{59, 46, 30, 23, 27, 40, 52, 35\} $ and $29$. The median of the data is :
AnswerArranging the given data in ascending order, we get
$\{23, 27, 29, 30, 35, 40, 46, 52, 59\}$
Here $, n = 9,$ which is odd.
$\therefore$ Median $=\Big(\frac{\text{n+1}}{2}\Big)^\text{th} \text{term}$
$=\Big(\frac{9+1}{2}\Big)^\text{th}\text{term}$
$=\big(\frac{10}{2}\big)^\text{th}\text{term}$
$= 5^\text{th} \text{ term}$
$=35$
View full question & answer→MCQ 1341 Mark
Look at the cumulative frequency distribution table given below :
|
Monthly income
|
Number of families
|
|
More that $Rs. 10000$
|
$100$ |
|
More that $Rs. 14000$
|
$85$ |
|
More that $Rs. 18000$
|
$69$ |
|
More that $Rs. 20000$
|
$50$ |
|
More that $Rs. 25000$
|
$37$ |
|
More that $Rs. 30000$
|
$15$ |
Number of families having income range $Rs. 20000$ to $Rs. 25000$ is: AnswerNumber of families having income more than $Rs. 20000 = 50$
Number of families having income more than $Rs. 25000 = 37$
Hence, number of families having Income range $20000$ to $25000 = 50 - 37 = 13$
View full question & answer→MCQ 1351 Mark
Choose the correct answer from the given four options : For the following distribution :
| Class |
$0-5$ |
$5-10$ |
$10-15$ |
$15-20$ |
$20-25$ |
| Frequency |
$10$ |
$15$ |
$12$ |
$20$ |
$9$ |
the sum of lower limits of the median class and modal class is : AnswerHere,
| Class |
Frequency |
Cumulative frequency |
| $0-5$ |
$10$ |
$10$ |
| $5-10$ |
$15$ |
$25$ |
| $10-15$ |
$12$ |
$37$ |
| $15-20$ |
$20$ |
$57$ |
| $20-25$ |
$9$ |
$66$ |
Now, $\frac{\text{N}}{2}=\frac{66}{2}=33,$ which lies in the interval $10-15.$
Therefore, lower limit of the median class is $10.$
The highest frequency is $20,$ which lies in the interval $15-20$.
Therefore, lower limit of modal class is $15.$
Hence, required sum is $10 + 15 = 25.$ View full question & answer→MCQ 1361 Mark
The arithmetic mean of $\{1, 2, 3, 4, ……………….., n\}$ is :
AnswerCorrect option: C. $\frac{\text{n}+1}{2}$
According to questions,
Arithmatic Mean $=\frac{1+2+3+....+\text{n}}{\text{n}}$
$=\frac{\frac{\text{n}\text{(n+1)}}{2}}{2}$
$=\frac{\text{n+1}}{2}$
View full question & answer→MCQ 1371 Mark
The measure of central tendency that can be obtained graphically is $............$
AnswerThe measure of central tendency that can be obtained graphically is Median.
The median is less affected by outliers and skewed data distribution
i.e. when the distribution is not symmetrical.
View full question & answer→MCQ 1381 Mark
For the following distribution :
|
Mark
|
$60-70$ |
$70-80$ |
$80-90$ |
$90-100$ |
$100-110$ |
|
Frequency
|
$10$ |
$15$ |
$12$ |
$20$ |
$9$ |
The sum of lower limits of the median class and modal class is : Answer
|
Mark
|
$60-70$ |
$70-80$ |
$80-90$ |
$90-100$ |
$100-110$ |
|
Frequency
|
$10$ |
$15$ |
$12$ |
$20$ |
$9$ |
|
Cumulative Frequency
|
$10$ |
$25$ |
$37$ |
$57$ |
$66$ |
Here $N = 66$
$\Rightarrow\frac{\text{N}}{2}=33$
$\therefore$ The median class is $80-90$ and Modal class is $90-100$
Sum of lower limits of Median class and Modal class $= 80 + 90 = 170$ View full question & answer→MCQ 1391 Mark
For a symmetrical frequency distribution, we have :
- A
Mean $ < $ mode $ < $ median
- B
Mean $ > $ mode $ > $ median
- ✓
Mean $=$ mode $=$ median
- D
Mode $=\frac{1}{2} \ ($mean $+$ median$)$
AnswerCorrect option: C. Mean $=$ mode $=$ median
For a symmetrical distribution, we have
Mean $=$ mode $=$ median
View full question & answer→MCQ 1401 Mark
In the formula, $\bar{\text{x}}=\Big\{\text{A}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}\Big\}$ for finding the mean of the grouped data, the $\mathrm{d}_{\mathrm{i}}$ 's are the deviations from $A$ of:
- A
Lower limits of the classes
- B
Upper limits of the classes
- ✓
- D
Answer$\mathrm{d}_{\mathrm{i}}$ 's are the deviations from $A$ of midpoints of the classes
View full question & answer→MCQ 1411 Mark
Consider the frequency distribution of the heights of $60$ students of a class :
| Height $($in $\ cm)$ |
No. of Students |
Cumulative Frequency |
| $150-155$ |
$16$ |
$16$ |
| $155-160$ |
$12$ |
$28$ |
| $160-165$ |
$9$ |
$37$ |
| $165-170$ |
$7$ |
$44$ |
| $170-175$ |
$10$ |
$54$ |
| $175-180$ |
$6$ |
$60$ |
The sum of the lower limit of the modal class and the upper limit of the median class is : AnswerClass having maximum frequency is the modal class.
Hence, modal class : $150- 155.$
$\therefore$ Lower Iimit of the modal class $= 150$
$\text{N}=60$
$\Rightarrow\frac{\text{N}}{2}=30$
The cumulative frequency just greater than $30$ is $37.$
Hence, the median class is $160-165.$
$\therefore$ Upper limit of the median class $= 165$
Required sum $= 150 + 165 = 315$
View full question & answer→MCQ 1421 Mark
The mean of $20$ numbers is zero. of them, at the most, how many may be greater than zero?
AnswerMean of $20$ numbers $= 0$
Hence, sum of $20$ numbers $= 0 \times 20 = 0$
Now, the mean can be zero if
Sum of $10$ numbers is $(S)$ and the sum of remaining $10$ numbers is $(-S),$
Sum of $11$ numbers is $(S)$ and the sum of remaining $9$ numbers is $(-S),$
Sum of $19$ nun'bers is $(S)$ and the $20^{th}$ number is $(-S),$ then their sum is zero.
So, at the most $19$ numbers can be greater than zero.
View full question & answer→MCQ 1431 Mark
The empirical relationship between the three measures of central tendencies is :
- ✓
Mode $= 3$ median $- 2$ mean
- B
$3$ mean $=$ median $+\ 2$ mode
- C
$3 $ mode $=$ mean $+\ 2$ median
- D
AnswerCorrect option: A. Mode $= 3$ median $- 2$ mean
The empirical relationship between the three measures of central tendencies is $3$ Median $=$ Mode $+\ 2$ Mean.
The relationship is as per observation.
A distribution in which the values of mean, median and mode coincide
$($i.e. mean $=$ median $=$ mode$)$ is called symmetrical distribution.
conversely, when the values of mean, median, mode are not equal, the distribution is called asymmetrical or skewed.
Knowing any two values, the third can be computed by this formula
$3$ median $= 2$ mean $+$ mode
$2$ mean $= 3$ median $-$ mode
mean $=\frac{1}{2}\ {3}$ median $-$ mode
View full question & answer→MCQ 1441 Mark
$2($Median $−$ Mean$) =$ Mode $−$ Mean
Answer$2($Median $−$ Mean$) =$ Mode $−$ Mean.
$2($Median$) =$ Mode $+$ Mean.
This essentially means that the statement is saying that the Median is the average of Mode and Mean in the given data, which is not correct.
View full question & answer→MCQ 1451 Mark
The percentage of marks obtained by $100$ students in an examination are as follows :
|
Mark
|
$30-35$ |
$35-40$ |
$40-45$ |
$45-50$ |
$50-55$ |
$55-60$ |
$60-65$ |
|
Frequency
|
$10$ |
$15$ |
$18$ |
$22$ |
$23$ |
$8$ |
$4$ |
The median class is : - A
$35-40$
- ✓
$45-50$
- C
$40-45$
- D
$50-55$
AnswerCorrect option: B. $45-50$
|
Classes
|
$30-35$ |
$35-40$ |
$40-45$ |
$45-50$ |
$50-55$ |
$55-60$ |
$60-65$ |
|
Frequency
|
$10$ |
$15$ |
$18$ |
$22$ |
$23$ |
$8$ |
$4$ |
|
Cumulative Frequency
|
$10$ |
$25$ |
$43$ |
$65$ |
$88$ |
$96$ |
$100$ |
Here $N = 100$
$\Rightarrow \frac{\text{n}}{2}=50$
$\therefore ,$ median class is $45–50.$ View full question & answer→MCQ 1461 Mark
The algebraic sum of the deviations of a frequency distribution from its mean is :
- A
- B
- ✓
$0.$
- D
A non $-$ zero number.
AnswerThe algebraic sum of the deviations of a frequency distribution from its mean is zero
Let $x_1, x_2, x_3, \ldots . . x_n$ are observations and $\overline{\text{X}}$ is the mean
$\therefore (\bar{\text{x}}-\text{x}_1)+(\bar{\text{x}}-\text{x}_2)+(\bar{\text{x}}-\text{x}_3)+ ....(\bar{\text{x}}-\text{x}_\text{n})$
$=\text{n}\bar{\text{x}}-(\text{x}_1+\text{x}_2+\text{x}_3+.....\text{x}_\text{n})$
$=\text{n}\bar{\text{x}}-\text{n}\bar{\text{x}}=0$
View full question & answer→MCQ 1471 Mark
Which one of the following is not a measure of central tendency?
AnswerMean, Median and Mode are the most common measures of central tendency.
These may be considered depending on the type of data and data distribution.
Variance measures how far the data set is spread out and is not a measure of central tendency.
View full question & answer→MCQ 1481 Mark
In a data, if $\mathrm{l}=40, \mathrm{~h}=15, \mathrm{f}_1=7, \mathrm{f}_0=3, \mathrm{f}_2=6$, then the mode is :
Answer$\text{Mode}=\text{l}+\Big(\frac{\text{f}_1-\text{f}_0}{2\text{f}_1-\text{f}_0-\text{f}_2}\Big)\times\text{h}$
$=40+\frac{7-3}{7\times2-3-6}\times 15$
$=40+\frac{4}{5}\times 15$
$=40+12$
$=52$
View full question & answer→MCQ 1491 Mark
The median of first $10$ prime numbers is :
AnswerFirst $10$ prime numbers are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\}$
Here $n = 10$
$\therefore\text{Median}=\frac{1}{2}\Big[\frac{\text{n}}{2}\text{th}+\Big(\frac{\text{n}}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}\Big[\frac{10}{2}\text{th}+\Big(\frac{10}{2}+1\Big)\text{th}\Big]\text{term}$
$=\frac{1}{2}[5\text{th}+6\text{th}]\text{term}$
$=\frac{1}{2}[11+13]=\frac{1}{2}\times24=12$
View full question & answer→MCQ 1501 Mark
One of the method of determining mode is:
Answerc. Mode = 3 Median - 2 Mean
Solution:
It is well known fact that for moderately skewed matrix:
Mode = 3 Median - 2 Mean
View full question & answer→