MCQ Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 7

The line graph represent the net sales rate of Apple company from the year

$2003-2013$ . Find the increase in sales rate during

$2005-2007$ .

0%
$16,000$ millions
0%
$8,000$ millions
0%
$26,000$ millions
0%
$32,000$ millions

Explanation

The difference between the two sales rate of the year

$2005-2007$ is:

$32,000 - 16,000 = 16,000$
So, the increase in sales rate from the year

$2005-2007 = 16,000$ millions

According to the graph above, in which year was the ratio of the number of students enrolled at School

$B$ to the number of students enrolled at School

$A$ the greatest?

0%
$1990$
0%
$1991$
0%
$1992$
0%
$1993$
0%
$1994$

Explanation

As per given graph:

In year

$1990$ number of students enrolled at school

$B$ is

$1200$ and school

$A$ is

$800$

Then ratio of

$B$ to

$A$

$=$

$\dfrac{1200}{800}=\dfrac{3}{2}$

In year

$1991$ number of students enrolled at school

$B$ is

$800$ and school

$A$ is

$1600$

Then ratio of

$B$ to

$A$

$=$

$\dfrac{800}{1600}=\dfrac{1}{2}$

In year

$1992$ number of students enrolled at school

$B$ is

$2000$ and school

$A$ is

$1200$

Then ratio of

$B$ to

$A$

$=$

$\dfrac{2000}{1200}=\dfrac{5}{3}$

In year

$1993$ number of students enrolled at school

$B$ is

$1600$ and school

$A$ is

$1600$

Then ratio of

$B$ to

$A$

$=$

$\dfrac{1600}{1600}=\dfrac{1}{1}$

In year

$1994$ number of students enrolled at school

$B$ is

$1600$ and school

$A$ is

$800$

Then ratio of

$B$ to

$A$

$=$

$\dfrac{1600}{800}=\dfrac{2}{1}$

Then in year

$1994$ the ratio is greatest

So, answer is

$(E)$ ,

$1994$ is correct.

The mean of a finite set X of numbers is

$14$ , the median of this set of numbers is

$12$ , and the standard deviation is

$1.8$ . A new set Y is formed by multiplying each member of the set S by

$3$ . Determine the correct statements w.r.t. set Y:
I. The mean of the numbers is set Y is

$42$
II. The median of the numbers in set Y is

$36$
III. The standard deviation of the numbers is set Y is

$5.4$

0%
I only
0%
II only
0%
I and II only
0%
I and III only
0%
I, II, and III

Explanation

For set

$x$

Mean

$=14$

Median

$=12$

S.D.

$=\sqrt { \sigma }$

$=1.8$ .

For set

$y$

Mean

$=14\times 3= 42$ .

Median

$=12\times 3=36$ .

$=1.8\times 3= 5.4$ .

Hence all three are correct.

How many watches were sold in all 5 months?

0%
160
0%
180
0%
175
0%
170

Explanation

Total Sales= Addition of sales individually in each month.

Total Sale

$=20+50+20+60+30=180$

Which year did the same number of boys and girls attend the conference?

0%
1995
0%
1996
0%
1997
0%
1998
0%
None

Grades for the test on proofs did not go as well as the teacher had hoped. The mean grade was 68, the median grade was 64, and the standard deviation wasThe teacher curves the score by raising each score by a total of 7 points. Which of the following statements is true?
I. The new mean is 75.
II. The new median is 71.
III. The new standard deviation is 7.

0%
I only
0%
III only
0%
I and II only
0%
I, II, and III
0%
None of the statements are true

Explanation

If each observation is increased by

$7$ then the mean will also be increased by

$7$ .

New Mean = Old Mean

$+7=68+7=75$

If each observation is increased by

$7$ then the median will also be increased by

$7$ .

New median = Old median

$+7=64+7=71$

If each observation is increased/decreased by any quantity the standard deviation will remain same.

New standard deviation= old standard deviation

$=12$

The figure above shows the cost of joining and buying music from a music subscription service. What does the y-intercept of the line most likely represent?

0%
The cost per song
0%
The cost to join the service
0%
The cost of buying 20 songs
0%
The cost of 20 subscriptions to the service

Explanation

When no. of songs are 0 then cost is 20

$. This means the cost of joining the service is 20$ .

Hence, the y-intercept (i.e. 20) of the line represent the cost to join the service.

So, the answer is option (B).

2063760_493896_ans_8524580c8694437295258b86168e7a40.png

In a final buzzer round of a competition, contestants were asked to name as many states that begin with the letter

$K$ , by pressing the buzzer as early as possible., The bar graph shows the number of states the contestants were able to name. Find the number of contestants who participated in final buzzer round.

0%
$6$
0%
$8$
0%
$14$
0%
$20$

Explanation

Arranging the data in tabular form gives us

No of states $\rightarrow$	1	2	3	4	5	6	7	8	Total
No of candidates who correctly answered them $\rightarrow$	1	1	5	6	4	0	2	1	20

Hence, the total number of participants were

$20$ .

A random variable

$X$ has the probability distribution given below. Its variance is

X	1	2	3	4	5
P(X=x)	k	2k	3k	2k	k

0%
$\dfrac{16}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{5}{3}$
0%
$\dfrac{10}{3}$

Explanation

We know that

$\sum P(x) = 1$

$k+2k+3k+2k+k=1$

$9k=1$

$k=\frac{1}{9}$

Now, var

$\sigma^2 = \sum xi^2P(xi)-(\sum\,xiP(xi))^2$

$=1(k)+2^2(2k)+3^2(3k)+4^2(2k)+5^2(k)-[1(k)+2(2k)+3(3k)+4(2k)+5(k)]^2$

$=93k-(27k)^2$

$=93(\cfrac{1}{9})-[27(\frac{1}{9})]^2$

$=\cfrac{31}{3}-3^2$

$=\cfrac{31-27}{3}$

$=\cfrac{4}{3}$

What are the advantages of squaring a difference for calculating variance and standard deviation?

0%
Squaring makes each term positive so that values above the mean do not cancel below the mean.
0%
Squaring adds more weight to the larger differences, and in many cases this extra weight is appropriate since points further from the mean may be more significant.
0%
It complicates the calculations
0%
All are incorrect

Explanation

Since,

$\sigma_x=\sqrt{\cfrac{\sum (x_i-\bar x)^2}{N}}$

So, we can say that opion

$A$ and

$B$ are correct.

In how many of the given years was the production of fertilizers more than the average production of the given years?

0%
1
0%
2
0%
3
0%
4

Explanation

Average production (in 10000 tonnes) over the given years

$= \dfrac{1}{8} (25 + 40 + 60 + 45 + 65 + 50 + 75 + 80) = 55$ .

$\therefore$ The productions during the years 1997, 1999, 2001 and 2002 are more than the average production.

For how many companies has there been no decrease in production in any year from the previous year?

0%
One
0%
Two
0%
Three
0%
Four

In which pair of years was the number of candidates qualified, the same?

0%
1995 and 1997
0%
1995 and 2000
0%
1998 and 1999
0%
Data inadequate

The difference in the sales of cellular phones for the years

$1997$ and

$1999$ is?

0%
$500$ units
0%
$1,000$ units
0%
$5,000$ units
0%
$18,000$ units

Explanation

The given graph shows the sale of cellular phones in the years

$1997, 1998, 1999, 2000,2001$ and

$2002$ .

In the year

$1997$ , the sale of cellular phones is

$48000$ .

In the year

$1999$ , the sale of cellular phones is

$30000$ .

Therefore, the difference in the sales of cellular phones for the year

$1997$ and

$1999$ is,

$48000-30000=18000$

Hence, the difference in the sales of cellular phones for the year

$1997$ and

$1999$ is

$18,000$ units.

Thus, option

$D$ is correct.

Suppose a researcher is concerned with a nominal scale that identifies users versus nonusers of bank credit cards. The measure of central tendency appropriate to this scale is the

0%
Mean
0%
Mode
0%
Median
0%
Average

The average production for five years was maximum for which company?

0%
X
0%
Y
0%
Z
0%
X and Z both

Explanation

Average production (in lakh tons) in five years for the three companies are:
For Company X

$= \left [ \dfrac{1}{5} \times ( 30 +45 + 25 + 50 + 40) \right] = \dfrac{190}{5} = 38$
For Company Y

$= \left [ \dfrac{1}{5} \times ( 25 +35 + 35 + 40 + 50) \right] = \dfrac{185}{5} = 37$
For Company Z

$= \left [ \dfrac{1}{5} \times ( 35 +40 + 45 + 35 + 35) \right] = \dfrac{190}{5} = 38$

$\therefore$ Average production of five years is maximum for both the Companies X and Z.

What was the percentage increase in imports from 1997 to 1998 ?

0%
72
0%
56
0%
28
0%
Data inadequate

The degree to which numerical data tend to spread about value is called

0%
mean
0%
variation
0%
median
0%
mode

Explanation

Variance gives the degree to which the numerical data tend to spread about the value.

Variance is given by

$\dfrac{\sum x^2}{n}-(\dfrac{\sum x}{n})^2$

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of

$10$ to each of the students. Which of the following statistical measures will not change even after the grace marks were given.

0%
Median
0%
Mode
0%
Variance
0%
Mean

$x$ is increased by

$k$ then

$\sigma$ changes to

0%
$k+\sigma$
0%
$k\sigma$
0%
$k\sqrt{\sigma}$
0%
remains unchanged

Explanation

By the properties of standard deviation,

$\sigma(x+k)=\sigma(x)$

Therefore, by the addition of the constant, the standard deviation remains unchanged.

Variance of the first

$11$ natural numbers is:

0%
$\sqrt{5}$
0%
$\sqrt{10}$
0%
$5\sqrt{2}$
0%
$10$

Explanation

Sum of first

$n$ natural numbers is

$\dfrac {n^2-1}{12}$

$=\dfrac {11^2-1}{12}$

$=\dfrac {120}{12}$

$=10$

Thus, variance is

$\sqrt {\dfrac {n^2-1}{12}}$

$=\sqrt {10}$ .

$x$ is multiplied by

$k$ then

$\sigma$ changes to

0%
$k^2\sigma$
0%
$k\sigma$
0%
$k\sqrt{\sigma}$
0%
$k(\sigma)^2$

Explanation

By the properties of standard deviation,

$\sigma(kx)=|k|\sigma(x)$

Therefore, by multiplying the

$x$ by

$k$ , SD changes to

$k\sigma$

Standard Deviation of n observations

$a_1, a_2, a_3 .....a_n$ is

$\sigma$ Then the standard deviation of the observations

$\lambda a_1, \lambda a_2, ..., \lambda a_n$ is

0%
$\lambda \sigma$
0%
$-\lambda \sigma$
0%
$|\lambda| \sigma$
0%
${\lambda}^n \sigma$

Explanation

Let the mean of original data is

$\mu$

Then mean of observation if all the terms are multiplied by the constant

$\lambda$ will be

$\lambda \mu$

$\sigma$ be the standard deviation of original observation , then

$\sigma=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n (x_i-\mu)^2 }$

Now let the standard deviation of new observation is

$\sigma'$ , then

$\sigma'=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n (\lambda x_i-\lambda\mu)^2 }$

$=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n \lambda^2( x_i-\mu)^2 }$

$=|\lambda|\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n( x_i-\mu)^2 }=|\lambda |\sigma$

Note that: standard deviation can never be negative

The formula to find variance is

0%
${\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)}$
0%
${\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$
0%
${\dfrac{\sum x^2}{n}-(\overline x)^2}$
0%
${\dfrac{\sum x}{n}-\left(\dfrac{\sum x}{n}\right)^2}$

Explanation

Variance gives the degree to which the numerical data tend to spread about the value.

Variance is given by

$\dfrac{\sum x^2}{n}-(\dfrac{\sum x}{n})^2$

But mean

$\bar x=\dfrac{\sum x}{n}$

Therefore, variance is

$\dfrac{\sum x^2}{n}-(\bar x)^2$

The standard deviation of

$a,a+d,a+2d,...,a+{2}{n}d$ is

0%
$nd$
0%
${n}^{2}d$
0%
$\sqrt { \cfrac { n(n+1) }{ 3 } } d\quad$
0%
$\sqrt { \cfrac { n(n+3) }{ 3 } } d\quad$

Explanation

Series is

$a, a+d, a+2d,\cdots, a+2nd$

Total number of terms:

$N = 2n+1$

Sum of the series:

$S = \cfrac{2n+1}2(a+a+2nd) = (2n+1)(a+nd)$

Mean of the data:

$\bar{x} = \cfrac SN = \cfrac{(2n+1)(a+nd)}{(2n+1)} = a+nd$

Standard Deviation:

$s.d. = \sqrt{\cfrac{\displaystyle\sum_{i=1}^{2n+1} (x_i - \bar x)^2}{N}}$

$s.d. = \sqrt{\cfrac{2(1^2+2^2+\cdots+n^2)}{(2n+1)} = \cfrac{2\times\dfrac{n(n+1)(2n+1)}{6}}{(2n+1)}d^2}$

$s.d. = \sqrt{\cfrac{n(n+1)}{3}}d$

If the variance of a data is

$12.25$ , then the standard deviation is

0%
$3.5$
0%
$3$
0%
$2.5$
0%
$3.25$

Explanation

$S.D=\sqrt{variance}$

$\Rightarrow S.D=\sqrt{12.25}$

$\Rightarrow S.D=3.5$
Option A is correct.

Find the variance of the following distribution.

Class interval	$20-24$	$25-29$	$30-34$	$35-39$	$40-44$	$45-49$
Frequency	$15$	$25$	$28$	$12$	$12$	$8$

0%
$416$
0%
$46$
0%
$16$
0%
None of these

Explanation

$Class$	$Mid-point$ $(x)$	$Frequency$ $(f)$	$xf$	$(x-\overline{x})$	$(x-\overline{x})^2$	$(x-\overline{x})^2f$
$20-24$	$22$	$15$	$330$	$-10.25$	$105.0625$	$1575.9375$
$25-29$	$27$	$25$	$675$	$-5.25$	$27.5625$	$689.0625$
$30-34$	$32$	$28$	$896$	$-0.25$	$0.0625$	$1.75$
$35-39$	$37$	$12$	$444$	$4.75$	$22.5625$	$270.75$
$40-44$	$42$	$12$	$504$	$9.75$	$95.0625$	$1140.75$
$45-49$	$47$	$8$	$376$	$14.75$	$217.5625$	$1740.5$
		$\sum f=100$	$\sum xf=3225$			$\sum(x-\overline{x})^2f=5418.75$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum xf}{\sum f}=\dfrac{3225}{100}=32.25$

$\Rightarrow$

$Variance(\sigma^2)=\dfrac{\sum(x-\overline{x})^2f}{\sum f}=\dfrac{5418.75}{100}=54.18$

The mean and the variance of

$10$ observations are given to be

$4$ and

$2$ respectively. If every observation is multiplied by

$2$ , the mean and the variance of the new series will be respectively.

0%
$8$ and $20$
0%
$8$ and $4$
0%
$8$ and $8$
0%
$80$ and $40$

Explanation

Mean is summation of all observations divided by number of observation and hence, would be multiplied by

$2$ due to linearity. Hence, the new mean is

$8$ .

Variance is sum of squares of each observation subtracted by the mean. Hence, due to squared dependence, it will be quadrupled to get

$8$ .

Hence, both mean and variance are

$8$ .

Calculate the standard deviation of the following data.

x	$3$	$8$	$13$	$18$	$23$
f	$7$	$10$	$15$	$10$	$8$

0%
$6.32$ .
0%
$2.32$ .
0%
$3.32$ .
0%
None of these

Explanation

$x$	$f$	$xf$	$(x-\overline{x})$	$(x-\overline{x})^2$	$(x-\overline{x})^2f$
$3$	$7$	$21$	$-10.2$	$104.04$	$728.28$
$8$	$10$	$80$	$-5.2$	$27.04$	$270.4$
$13$	$15$	$195$	$-0.2$	$0.04$	$0.6$
$18$	$10$	$180$	$4.8$	$23.04$	$230.4$
$23$	$8$	$184$	$9.8$	$96.04$	$768.32$
	$\sum f=50$	$\sum xf=660$			$\sum(x-\overline{x})^2=1998$

$\Rightarrow$

$\overline{x}=\dfrac{\sum xf}{\sum f}=\dfrac{660}{50}=13.2$

Now, calculate standard deviation :

$S=\sqrt{\dfrac{(x-\overline{x})^2}{\sum f}}=\sqrt{\dfrac{1998}{50}}=\sqrt{39.96}=6.32$

What is the combined standard deviation of all 250 items ?

0%
$7.45$
0%
$7.35$
0%
$7.98$
0%
$7.73$

Explanation

Here

$n_1=100,\bar{x}_1=50, s_1=5,n_2=150,\bar{x}_2=40,s_2=6$

$\therefore \sigma=\sqrt{\dfrac{n_1s_1^2+n_2s_2^2}{n_1+n_2}+\dfrac{n_1n_2(\bar{x}_1-\bar{x}_2)^2}{(n_1+n_2)^2}}$

$=\sqrt{\dfrac{100(5)^2+150(6)^2}{100+150}+\dfrac{(100)(150)(50-40)^2}{(100+150)^2}}$

$=\sqrt{\dfrac{158}{5}+24}$

$=\sqrt{31.6+24}$

$=\sqrt{55.6}$

$=7.45$

$\therefore \sigma=7.45$

Let X denote the number of scores which exceed 4 in 1 toss of a symmetrical die. Consider the following statements :
The arithmetic mean of X is
The standard deviation of X is
Which of the above statements is/are correct ?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

The only possible favorable outcomes are

$5$ and

$6$ . Hence, the arithmetic mean of

$X$ is 5.5.

The standard deviation of

$X=\dfrac{1}{2}((6-5.5)^2+(5-55)^2)=0.25$

The variance of numbers

$x_1, x_2, x_3,..... x_n$ is V. Consider the following statements :
If every

$x_1$ , is increased by 2, the variance of the new set of numbers is V.
If the numbers

$x_1$ is squared, the variance of the new set is

$V^2$ .
Which of the following statements is/are correct ?

0%
1 only
0%
2 only
0%
Both 1 and 2
0%
Neither 1 nor 2

Explanation

In case 1, a deterministic value of 2 units is being added to every

$x_i$ and hence, the variance of the set remains unchanged as

$V$ . New variance would be

$\dfrac{1}{n}\sum\left(x_i+2-\left(\sum\dfrac{x_i+2}{n}\right)\right)=\dfrac{1}{n}\sum\left(x_i-\sum\dfrac{x_i}{n}\right)=V$

In case 2, each

$x_i$ is squared and hence, the mean becomes

$\dfrac{1}{n}\sum x_i^2$ . Hence, the new variance is

$\dfrac{1}{n}\sum\left( x_i^2-\dfrac{1}{n}\sum x_i^2\right)^2$

and not

$V^2=\dfrac{1}{n^2}\sum\left( x_i-\dfrac{1}{n}\sum x_i\right)^4$ .

Calculate the standard deviation of the first

$13$ natural numbers.

0%
$3.74$ .
0%
$2.74$ .
0%
$3.48$ .
0%
None of these

Explanation

$\Rightarrow$

$1st$

$13$ natural numbers are

$1,2,3,4,5,6,7,8,9,10,11,12,13.$

$\sum x=\dfrac{n(n+1)}{2}$

$=\dfrac{13(13+1)}{2}$

$=\dfrac{13\times 14}{2}$

$\therefore$

$\sum x=91$

Now,

$\sum x^2=\dfrac{n(n+1)(2n+1)}{6}$

$=\dfrac{13\times 14\times 27}{6}$

$\therefore$

$\sum x^2=819$

$\Rightarrow$

$SD=\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$

$=\sqrt{\dfrac{819}{13}-\left(\dfrac{91}{13}\right)^2}$

$=\sqrt{63-49}$

$=\sqrt{14}$

$=3.74$

The variance of first '

$n$ ' natural number is

0%
$\dfrac { { n }^{ 2 }+1 }{ 12 }$
0%
$\dfrac { { n }^{ 2 }-1 }{ 12 }$
0%
$\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 }$
0%
None of these

Explanation

By using formula of variance, we have

${ \sigma }^{ 2 }=\dfrac { 1 }{ n } \displaystyle\sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } -{ \left( \dfrac { 1 }{ n } \displaystyle\sum _{ i=1 }^{ n }{ { x }_{ i } } \right) }^{ 2 }$

$=\dfrac { 1 }{ n } \left( { 1 }^{ 2 }+{ 2 }^{ 2 }+\cdots +{ n }^{ 2 } \right) -{ \left( \dfrac { 1 }{ n } \left( 1+2+\cdots +n \right) \right) }^{ 2 }$

$=\dfrac { 1 }{ n } \cdot \dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } -\dfrac { 1 }{ { n }^{ 2 } } \cdot { \left[ \dfrac { n\left( n+1 \right) }{ 2 } \right] }^{ 2 }$

$=\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 } -\dfrac { { \left( n+1 \right) }^{ 2 } }{ 4 }$

$=\dfrac { { n }^{ 2 }-1 }{ 12 }$

Calculate the standard deviation of the following data.

$10, 20, 15, 8, 3, 4$

0%
$5.97$
0%
$59.7$
0%
$4.97$
0%
None of these

Explanation

$\Rightarrow$ The given numbers are

$3,4,8,10,15,20$ .

$\Rightarrow$

$\overline{X}=\dfrac{3+4+8+10+15+20}{6}=10$

$X$	$X-\overline{X}$	$(X-\overline{X})^2$
$3$	$-7$	$49$
$4$	$-6$	$36$
$8$	$-2$	$4$
$10$	$0$	$0$
$15$	$5$	$25$
$20$	$10$	$100$
		$\sum (X-\overline{X})^2=214$

$\Rightarrow$

$S=\sqrt{\dfrac{(X-\overline{X})^2}{N}}$

$=\sqrt{\dfrac{214}{6}}$

$=\sqrt{35.66}$

$=5.97$

What is the ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years?

0%
2:3
0%
3:5
0%
4:5
0%
7:9

Explanation

Total sale of

$B2=75+65=140$

Total sale of

$B4=85+95=180$

Ratio of sales of

$B2$ and

$B4$

$=\dfrac{140}{180}=\dfrac{7}{9}$

So option

$D$ is correct.

If the mean of the numbers

$a,b,8,5,10$ is

$6$ and their variance is

$68$ , then

$ab$ is equal to

0%
$6$
0%
$7$
0%
$12$
0%
$14$
0%
$25$

Explanation

Given,

$\cfrac { a+b+8+5+10 }{ 5 } =6$

$\Rightarrow a+b+c+23=30$

$\Rightarrow a+b=7....(i)$
and

$\quad \cfrac { \sum { { \left( { x }_{ 1 }-\overline { x } \right) }^{ 2 } } }{ n } ={ \sigma }^{ 2 }\quad$

$\cfrac { { (a-6) }^{ 2 }+{ (b-6) }^{ 2 }+4+1+16 }{ 5 } =\cfrac { 34 }{ 5 } \quad$

$\Rightarrow { (a-6) }^{ 2 }+{ (b-6) }^{ 2 }+21=34\quad$

$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }-84+93=34\quad \left[ \because a+b=7 \right]$

$\quad \Rightarrow { a }^{ 2 }+{ b }^{ 2 }=25...(i)\quad$
From Eq.(i)

$\Rightarrow { (a+b) }^{ 2 }={ 7 }^{ 2 }$

$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }+2ab=49$

$\Rightarrow 25+2ab=49$ (From Eq.(ii))

$\Rightarrow 2ab=24\Rightarrow ab=12\quad$

For the data set below, which value is in the 75th percentile?

$1, 3, 3, 4, 6, 7, 7, 7, 8, 9, 9, 10, 12, 15, 16, 17$

0%
16
0%
9
0%
12
0%
15

Explanation

$1, 3, 3, 4, 6, 7, 7, 7, 8, 9, 9, 10, 12, 15, 16, 17$

Logic:- first quartile lies halfway between the lowest valve of the Median & the third Quartile lies halfway between the Median & the largest value.

$N=16$

Third quartile or

$75$ percentile

$=\dfrac{3(n+1)}{4}$ th term

$=\dfrac{3\times 17}{4}=12.75^{th}$ term

$=12^{th}$ term

$+0.75\times(13^{th}$ term

$-12^{th}$ term

$)$

$=10+0.75\times (12-10)$

$=10+0.75\times 2$

$=11.5\approx 12$ .

1154293_697667_ans_15a9daf0af844501a65e46105d0518f4.jpg

What is the total expenditure (Rs. in lakhs) of the company during the period January to July in the year 2003?

0%
$3,800$
0%
$3,950$
0%
$4,600$
0%
$5,350$

Explanation

Required expenditure of the company

$=600+500+500+650+500+600+600$

$3950$ lakh

If the variance of

$1, 2, 3, 4, 5, ..., x$ is

$10$ , then the value of

$x$ is

0%
$9$
0%
$13$
0%
$12$
0%
$10$
0%
$11$

Explanation

We know that, the variance of first

$x$ natural number is

$\text{Var} (x) = \dfrac {x^{2} - 1}{12} = 10$ ....(given)

$\Rightarrow x^{2} = 120 + 1 = 121$

$\Rightarrow x = 11$

For the information given below,Find quartile deviation and coefficient of quartile deviation

Maximum Load	Number of Cables
9.3-9.7	22
9.8-10.2	55
10.3-10.7	12
10.8-11.2	17
11.3-11.7	14
11.8-12.2	66
12.3-12.7	33
12.8-13.2	11

0%
$0.275,0.065$
0%
$0.495,0.045$
0%
$0.415,0.055$
0%
$0.225,0.012$

Explanation

$Maximum\,load$	$No.\,of\,cables$	$Class\,boundaries$	$Cumulative\,frequency$
$9.3-9.7$	$2$	$9.25-9.75$	$2$
$9.8-10.2$	$5$	$9.75-10.25$	$7$
$10.3-10.7$	$12$	$10.25-10.75$	$19$
$10.8-11.2$	$17$	$10.75-11.25$	$36$
$11.3-11.7$	$14$	$11.25-11.75$	$50$
$11.8-12.2$	$6$	$11.75-12.25$	$56$
$12.3-12.7$	$3$	$12.25-12.75$	$59$
$12.8-13.2$	$1$	$13.25$	$60$

$Q_1:$

Value of

$\dfrac{n}{4}^{th}=\dfrac{60}{4}=15^{th}$

$\therefore$

$Q_1$ lies in class

$10.25-10.75$

Where,

$l=10.25,\,h=0.5,\,f=12,\,cf=7$

$Q_1=l+\dfrac{h}{f}\left(\dfrac{n}{4}-c.f.\right)$

$=10.25+\dfrac{0.5}{12}(15-7)$

$=10.25+0.33$

$=10.58$

$Q_3:$

Value of

$\dfrac{3n}{4}^{th}item=\dfrac{3\times 60}{4}=45^{th}item$

$\therefore$

$Q_3$ lies in class

$11.25-11.75$

Here,

$l=11.25,\,h=0.5,\,f=14,\,\dfrac{3n}{4}=45$ and

$cf=36$

$Q_3=l+\dfrac{h}{f}\left(\dfrac{3n}{4}-c.f\right)$

$=11.25+\dfrac{0.5}{14}(45-36)$

$=11.25+0.32$

$=11.57$

$\Rightarrow$ Now, Quartile deviation

$=\dfrac{Q_3-Q_1}{2}$

$=\dfrac{11.57-10.58}{2}$

$=\dfrac{0.99}{2}$

$=0.495$

$\Rightarrow$ Coefficient of Quartile Deviation

$=\dfrac{Q_3-Q_1}{Q_3+Q_1}$

$=\dfrac{11.57-10.58}{11.57+10.58}$

$=\dfrac{0.99}{22.15}$

$=0.045$

The arithmetic mean of the observations 10,8,5,a,b is 6 and their variance 6.Then ab=

0%
6
0%
4
0%
3
0%
12

Explanation

$\dfrac{10+8+5+a+b}{5}=6,$

$\dfrac{1}{n}\sum \left ( x_i-\overline{x} \right )^{2}=6.8$

solving above two equations we get values of a and b.

Mean of

$10$ observations is

$50$ and their standard deviation is

$10$ . If each observation is subtracted by

$5$ and then divided by

$4$ , then the new mean and standard deviation are

0%
$22.45, 2.5$
0%
$11.25, 2.5$
0%
$11.5, 2.5$
0%
$11, 2.5$
0%
$11.75, 2.5$

Explanation

Given,mean of

$10$ observations

$= \dfrac {\displaystyle \sum_{i = 1}^{10} x_{i}}{10} = 50$

$\Rightarrow \displaystyle \sum_{i = 1}^{10} x_{i} = 500$ .... (i)
Therefore, new mean

$= \dfrac {\displaystyle \sum_{i = 1}^{10} x_{i} - 5\times 10}{4\times 10} = \dfrac {500 - 50}{4\times 10}$ [using Eq. (i)]

$= \dfrac {450}{4\times 10} = 11.25$
and standard deviation

$= \dfrac {\text{old}(SD)}{4} = \dfrac {10}{4} = 2.5$ .

What is the average monthly expenditure (Rs. in lakhs) for January to July during the year 2005?

0%
$658.3$
0%
$766.7$
0%
$764.2$
0%
$657.1$

Explanation

Required average

$=\cfrac { 700+750+850+850+600+750+850 }{ 7 }$

$=\cfrac { 5350 }{ 7 } =764.2$ lakh

The line graph show the number of cars Jatin sold over the past

$5$ weeks.
If he was paid

$Rs \ 20000$ for every car sold, how much was he paid over the past

$5$ weeks?