MCQ Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 10

Find variance of the following data.

Class interval	Frequency
$4-8$	$3$
$8-12$	$6$
$12-16$	$4$
$16-20$	$7$

0%
$19$
0%
$20$
0%
$21$
0%
$23$

Explanation

$C.I$	$x_i$	$f_i$	$f_ix_i$	$(x_i-\overline{x})^2$	$f_i(x_i-\overline{x})^2$
$4-8$	$6$	$3$	$18$	$49$	$147$
$8-12$	$10$	$6$	$60$	$9$	$54$
$12-16$	$14$	$4$	$56$	$1$	$4$
$16-20$	$18$	$7$	$126$	$25$	$175$
		$\sum f_i=20$	$\sum f_ix_i=260$	$\sum(x_i-\overline{x})^2=84$	$\sum f_i(x_i-\overline{x})^2=380$

$\Rightarrow$

$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{260}{20}=13$

$\Rightarrow$

$Variance(\sigma^2)\dfrac{\sum f_i(x_i-\overline{x})^2}{\sum f_i}=\dfrac{380}{20}=19$

$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right) = 9}$ and

$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$ , then the standard deviation of the

$9$ items

${x_1},{x_2},.....,{x_9}$ is

0%
$2$
0%
$3$
0%
$9$
0%
$4$

Explanation

S.D of

$xi-5$ is

$\sigma =\sqrt{\dfrac{\sum_{i=1}^{9}(xi-5)^2}{9}-\left [ \dfrac{\sum_{i=1}^{9}(xi-5)^2}{9} \right ]^2}$

$\sigma =\sqrt{5-1}=2$

Suppose a population

$A$ has

$100$ observations

$101,102,........,200$ and another population

$B$ has

$100$ observations

$151,152......,250$ . If

$V_{A}and V_{B}$ represent the variences of the two populations respectively, then

$V_{A}/ V_{B}$ is

0%
$1$
0%
$9/4$
0%
$4/9$
0%
$2/3$

If the sum of mean and variance of binomial distribution is 4.8 for 5 trials then

0%
mean is 3
0%
variance is 0.6
0%
mean is 3.2
0%
variance is 0.8

Find the standard deviation of 10 observation 111,211,311,....1011.

0%
100 $\sqrt { 3 }$
0%
250
0%
300
0%
50 $\sqrt { 3 }$

The mean deviation about the mean of the set of first

$n$ natural numbers when

$n$ is an odd number.

0%
$\dfrac{n^{2}-1}{4n}$
0%
$\dfrac{n}{4}$
0%
$\dfrac{n^{2}+1}{4n}$
0%
$\dfrac{n^{2}-1}{12}$

Explanation

$\overline { X } =\cfrac { \sum _{ i=1 }^{ n }{ i } }{ n } =\cfrac { n(n+1) }{ 2n } =\cfrac { n+1 }{ 2 }$

$\left| { d }_{ i } \right| =\left| r-\left( \cfrac { n+1 }{ 2 } \right) \right|$ $

$\sum { \left| { d }_{ i } \right| } =\sum _{ r=1 }^{ n }{ \left| r-\left( \cfrac { n+1 }{ 2 } \right) \right| } =\sum _{ r=1 }^{ \cfrac { n+1 }{ 2 } }{ \cfrac { n+1 }{ 2 } } =r+\sum _{ r=\cfrac { n+1 }{ 2 } }^{ n }{ r-\left( \cfrac { n+1 }{ 2 } \right) }$

$\Rightarrow \sum { = } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) -\cfrac { 1 }{ 2 } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+3 }{ 2 } \right) -\left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) +\cfrac { n+1 }{ 4 } \left( \cfrac { n+1 }{ 2 } +n \right) \quad$

$\Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } \right) -\cfrac { \left( n+1 \right) \left( n+3 \right) }{ 8 } \Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } -\cfrac { n+3 }{ 2 } \right)$

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

Let

$\sigma^{2}$ is variance of following frequency distribution

$x_{1}$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$f_{1}$	$1$	$0$	$1$	$7$	$9$	$4$	$1$	$1$	$1$

then

$\sigma^{2}$ is equal to

0%
$2.4$
0%
$2.5$
0%
$2.6$
0%
$2.7$

The mean and variance of a random variable having a binomial distribution are

$4$ and

$2$ respectively , then

$P(X=1)$ is

0%
$1/32$
0%
$1/16$
0%
$1/8$
0%
$1/4$

Let

$x_i$ represents the outcome on a fair die and

$f_i$ be the corresponding frequency. The variance for random variable

$x_i$ with following frequency distribution, is

$x_i$	1	2	3	4	5	6
$f_i$	1	2	3	4	5	6

0%
$2$
0%
$3$
0%
$\dfrac{28}{9}$
0%
$\dfrac{20}{9}$

The mean and the standard devition (s.d) of five observations are

$9$ and

$0$ , respecively. If one of the observations is changed such that the mean of the new set of five obervatons becomes

$10$ , then their s. d. is:

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

Explanation

Since the standard deviation before the change was $0$ , all the observation was the mean, or $9$ . Since one observation was changed and the new mean is $10$ , we have the following equation.

$\dfrac{9+9+9+9+x}{5}=10$

$\Rightarrow \dfrac{36+x}{5}=10$

$\Rightarrow 36+x=50$

$\Rightarrow x=14$

The changed observation is $14$ .

All the observations are $\left\{ 9,9,9,9,14 \right\}$

Since the mean is $10$ ,

The variance is

$\dfrac{\left[ {{\left( 9-10 \right)}^{2}}+{{\left( 9-10 \right)}^{2}}+{{\left( 9-10 \right)}^{2}}+{{\left( 9-10 \right)}^{2}}{{\left( 14-10 \right)}^{2}} \right]}{5}$

$\Rightarrow \dfrac{\left( 1+1+1+1+16 \right)}{5}$

$\Rightarrow 4$

Hence, the standard deviation is $\sqrt{4}=2$

Let

$x_{1},\ x_{2},...x_{100}$ are

$100$ above observation such that

$\displaystyle \sum { { x }_{ 1 }=0 } ,\ \displaystyle \sum _{ 1\le i<j\le 100 }^{ }{ \left| { x }_{ i }{ x }_{ j } \right| =80000 }$ & mean deviation from their mean is

$5$ , then their standard deviation, is-

0%
$10$
0%
$30$
0%
$40$
0%
$50$

For the observations

${ x }_{ 1, }{ x }_{ 2 },{ x }_{ 3 },........{ x }_{ 18, }$ it is given that

$\sum _{i =1 }^{ 18 }{ ({ x }_{ i }-8)=9 }$ and

$\sum _{ j=1}^{ 18 }{ { ({ x }_{ j}-8) }^{ 2 }=45 }$ then the standard deviation of these eighteen observations is

0%
$\cfrac { 3 }{ 2 }$
0%
5
0%
$\cfrac { 5 }{ \sqrt { 2 } }$
0%
$\sqrt { \cfrac { 81 }{ 34 } }$

If the mean deviation about the median of the numbers

$a,2a,....,50a$ is

$50$ , then

$|a|$ equals:-

0%
$4$
0%
$5$
0%
$2$
0%
$3$

Explanation

Given series:-

$a, 2a, 3a, ....., 50a$

Median

$= \cfrac{25a + 26a}{2} = 25.5 a$

Mean deviation about median

$= 50 \quad \left( \text{Given} \right)$

Mean deviation

$= \cfrac{\sum_{i = 1}^{50}{\left| {x}_{i} - 25.5a \right|}}{50}$

$\therefore \cfrac{24.5a + 23.5a + 22.5a + ..... + 23.5a + 24.5a}{50} = 50$

$\Rightarrow a + 3a + 5a + ..... + 47a + 49a = 2500$

$\Rightarrow \cfrac{25}{2} \left( 2a + \left( 25 - 1 \right) 2a \right) = 2500$

$\Rightarrow 2a \times 25 = 200$

$\Rightarrow a = \cfrac{200}{50} = 4$

The variance of first

$50$ even natural numbers is:-

0%
$833$
0%
$831$
0%
$438$
0%
$\dfrac {437}{4}$

$\displaystyle\sum _{ i=1 }^{ 18 }{ \left( { x }_{ i }-8 \right) } =153,\displaystyle\sum _{ i=1 }^{ 18 }{ \left( { x }_{ i }-8 \right) ^{ 2 } } =45$ then standard deviation of

$x_{1},x_{2},........,x_{n}$ is

0%
$4/9$
0%
$9/4$
0%
$3/2$
0%
$2/3$

Standard deviation of four observations

$-1, 0, 1$ and k is

$\sqrt{5}$ then k will be?

0%
$2\sqrt{6}$
0%
$1$
0%
$2$
0%
$\sqrt{6}$

Explanation

$\sigma^2=\dfrac{\displaystyle\sum x^2_i}{n}-\left(\dfrac{\displaystyle\sum x_i}{n}\right)^2$

$\Rightarrow 5=\dfrac{1+0+1+k^2}{4}-\left(\dfrac{-1+0+1+k}{4}\right)^2$

$\Rightarrow 5=\dfrac{k^2+2}{4}=\dfrac{k^2}{16}$

$\Rightarrow 80=4k^2+8-k^2$

$\Rightarrow 72=3k^2$

$\Rightarrow k=2\sqrt{6}$ .

A student scores the following marks in five test:

$45, 54, 41, 57, 43$ . His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six test is

0%
$\dfrac{10}{\sqrt{3}}$
0%
$\dfrac{100}{\sqrt{3}}$
0%
$\dfrac{130}{3}$
0%
$\dfrac{10}{3}$

Explanation

Let

$x$ be the

$6^{th}$ observation

$\Rightarrow 45 + 54 + 41 + 57 + 43 + x = 48 \times 6 = 288$

$\Rightarrow x = 48$
Variance

$= \left(\dfrac{\sum x_i^2}{6} - (\bar{x})^2\right)$

$\Rightarrow$ variance

$=\dfrac{14024}{6} - (48)^2$

$=\dfrac{100}{3}$

$\Rightarrow$ standard deviation

$=\dfrac{10}{\sqrt{3}}$

For a random variable

$X$ . If

$E(X)=5$ and

$V(X)=6$ , then

$E(X^{2})$ is equal to

0%
$19$
0%
$31$
0%
$61$
0%
$11$

Explanation

As we know that,

$V{\left( X \right)} = E{\left( {X}^{2} \right)} - {\left( E{\left( X \right)} \right)}^{2}$

$\therefore 6 = E{\left( {X}^{2} \right)} - {5}^{2}$

$\Rightarrow E{\left( {X}^{2} \right)} = 6 + 25 = 31$

If both the mean and the standard deviation of

$50$ observations

${ x }_{ 1 },{ x }_{ 2 },......,{ x }_{ 50 }$ are equal to

$16$ , then the mean of

${ \left( { x }_{ 1 }-4 \right) }^{ 2 },{ \left( { x }_{ 2 }-4 \right) }^{ 2 },.....{ \left( { x }_{ 50 }-4 \right) }^{ 2 }$ is:

0%
$525$
0%
$380$
0%
$480$
0%
$400$

Explanation

Mean

$\quad \left( \mu \right) =\cfrac { \sum { { x }_{ i } } }{ 50 } =16$
standard deviation

$\left( \sigma \right) =\sqrt { \cfrac { \sum { { x }_{ i }^{ 2 } } }{ 50 } -{ \left( \mu \right) }^{ 2 } } =16\Rightarrow \left( 256 \right) \times 2=\cfrac { \sum { { x }_{ i }^{ 2 } } }{ 50 }$
New mean

$=\cfrac { \sum { ({ x }_{ i }-4)^{ 2 } } }{ 50 } =\cfrac { \sum { { x }_{ i }^{ 2 } } +16\times 50-8\sum { { x }_{ i }^{ } } }{ 50 } =\left( 256 \right) \times 2=16-8\times 16=400$

If expected value in n Bernoulli trials is

$8$ and variance is

$4$ . If

$P(x\leq 2)=\dfrac{k}{2^{16}}$ then value of k is?

0%
$1$
0%
$137$
0%
$136$
0%
$120$

Explanation

Let number of trials be n and probability of success

$=$ p, probability of failure

$=$ q
Given

$np=8, npq=4$

$\Rightarrow q=\dfrac{1}{2}, p=\dfrac{1}{2}, n=16$ (as

$p+q=1$ )

$p(x\leq 2)=\dfrac{^{16}C_0 +{^{16}C_1}+{^{16}C_2}{2^{16}}}=\dfrac{1+16+120}{2^{16}}=\dfrac{137}{2^{16}}$
Hence

$(2)$ .

If the data

$x_1, x_2, .., x_{10}$ is such that the mean of first four of these is

$11$ , the mean of the remaining six is

$16$ and the sum of square of all of these is

$2,000$ ; then the standard deviation of this data is?

0%
$4$
0%
$2$
0%
$\sqrt{2}$
0%
$2\sqrt{2}$

Let the observations

$x_i(1\leq i \leq 10)$ satisfy the equations,

$\displaystyle\sum^{10}_{i=1}(x_i-5)=10$ and

$\displaystyle\sum^{10}_{i=1}(x_i-5)^2=40$ . If

$\mu$ and

$\lambda$ are the mean and the variance of the observations,

$x_1-3, x_2-3, ....., x_{10}-3$ , then the ordered pair

$(\mu, \lambda)$ is equal to?

0%
$(6, 6)$
0%
$(3, 6)$
0%
$(3, 3)$
0%
$(6, 3)$

Explanation

$\displaystyle\sum_{i = 1}^{10} (x_i - 5) = 10$

$\displaystyle\Rightarrow \sum_{i = 1}^{10} (x_i -50) = 10$

$\displaystyle\Rightarrow \sum_{i = 1}^{10} x_i = 60$

$\displaystyle\sum_{i = 1}^{10} (x_i - 5)^2 = 40$

$\displaystyle\Rightarrow \sum_{i = 1}^{10} (x_i^2 + 25 - 10 x_i) = 40$

$\displaystyle\Rightarrow \sum_{i = 1}^{10} x_i^2 + 250 - 10 \times 60 = 40$

$\displaystyle\sum_{i = 1}^{10} x_i^2 = 390$

$i = \dfrac{1}{N} \sum(x_i - 3)^2 = \left(\dfrac{1}{N} \sum(x_i - 3) \right)^2$

$= \dfrac{\sum (x_i^2 + 9 - 6 x_i)}{10} - \left(\dfrac{1}{10} (\sum x_i - 30)\right)^2$

Mean

$= \mu = \displaystyle\dfrac{\sum_{i = 1}^{10} (x_i - 3)}{N}$

$= \dfrac{\sum_{i = 1}^{10} x _i - 30}{10}$

$\{\sum_{i = 1}^{10} x_i = 60\}$

$= \dfrac{60 - 30}{10}$

$= \dfrac{30}{10}$

$= 3$

$\mu = 3, \, \lambda = 3$

$\therefore$

$\boxed{(\mu, \lambda)\equiv(3, 3)}....Answer$

The bar graph as shown in above figure represents the heights (in cm) of

$50$ students of Class

$XI$ of a particular school. Study the graph and answer the following questions:
State whether true or false:
The number of a student is the class is in the range of

$160-164$ cm is

$6$ .

0%
True
0%
False

Explanation

The given bar graph represents the height (in cm) of $50$ students of class $XI$ of a particular school.

We can prepare frequency table from given bar graph as follows:

Heights(in cm) No. of students

$140-144$ $7$

$145-149$ $11$

$150-154$ $17$

$155-159$ $9$

$160-164$ $6$

$165$ -

Total number of students $=50$ .

Here, the number of students belonging to

$160-164$ is

$6$ .

That is, the given statement is true and option

$A$ is correct.

The bar graph shown in Fig. represents the circulation of newspapers in 10 languages. Study the bar graph and answer the following questions:
State whether true or false:
The number of newspapers published in Telugu is more than those published in Tamil.

0%
True
0%
False

The mean of

$5$ observation is

$4.4$ and variance is

$8.24$ . If three of the five observations are

$1,2$ and

$6$ , then what are the other two observations ?

0%
$9,16$
0%
$9,4$
0%
$81,16$
0%
$81,4$

Explanation

Let the other two observations be x and y .

Therefore, the series is

$1, 2, 6,x,y$

Now Mean

$x = 4.4 = \dfrac{(1+2+ 6 +x + y) }{5}$

$22 = 9 + x + y$

Therefore

$x + y = 13$ ... (1)

Also

variance

$= 8.24 =\dfrac{1}{n}\sum ( x_i - x)^2$

$8.24 = \dfrac{1}{5}[ (3.4)^2 + (2.4)^2 + (1.6)^2 + x^2 +y^2 – 2(4.4) (x + y) + 2(4.4)^2]$

$41.20 = 11.56 + 5.76 + 2.56 + x^2 + y^2 – 8.8\times13 + 38.72$

Therefore

$x^2 + y^2 = 97$ …....... (2)

But from (1), we have

$x^2 + y^2 + 2xy = 169$ …......... (3)

From (2) and (3), we have

$2xy = 72$ ... (4)

Subtracting (4) from (2), we get

$x^2 + y^2 – 2 xy = 97 – 72$

i.e.

$(x– y)^2 = 25$

$x – y = \pm 5$ …..... (5)

So, from (1) and (5), we get

$x = 9, y = 4$ when

$x – y = 5$

$x = 4, y = 9$ when

$x – y = – 5$

Thus, the remaining observations are

$9,4$

The given bar graph represents the heights (in cm) of

$50$ students of Class

$XI$ of a particular school. Study the graph and answer the following question:
State true or false:
Number of students with maximum height (in cm) in the class is

$17$ .

0%
True
0%
False

Explanation

$\textbf{Step 1: Construct the frequency table.}$

$\text{The given bar graph represents the height (in cm) of }50$ $\text{students of class }XI$ $\text{of a particular school.}$

$\text{We can prepare frequency table from given bar graph as follows:}$

$\text{Heights(in cm)}$ $\text{No. of students}$

$140-144$ $7$

$145-149$ $11$

$150-154$ $17$

$155-159$ $9$

$160-164$ $6$

$\text{Total number of students }=50$ .

$\textbf{Step 2: Find number of students with maximum height.}$

$\text{Here, the maximum range of height is}$

$160-164 cm.$

$\text{And the number of students belonging to this group is }6$ .

$\text{Therefore, there are 6 students in the class in the range of maximum height of the class.}$

$\textbf{Hence, the given statement is false and option B is correct.}$

The mean and variance of

$20$ observations are found to be

$10$ and

$4$ , respectively. On rechecking, it was found that an observation

$9$ was incorrect and the correct observation was

$11$ . Then the correct variabce is:

0%
$3.98$
0%
$4.02$
0%
$4.01$
0%
$3.99$

Explanation

$\dfrac{\sum x_i}{20} = 10 \Rightarrow \sum x_i = 200$

$\dfrac{\sum x_{i}^2}{20} - 100 = 4$

$\sum x_i^2 = 104 \times 20$

$=2080$

Actual mean

$= \dfrac{200 - 9 + 11}{20}$

$= \dfrac{202}{20}$

Variance

$= \dfrac{2080 - 81 + 121}{20} - \left(\dfrac{202}{20} \right)^2$

$= \dfrac{2120}{20} - (10.1)^2$

$= 106 - 102.01$

$= 3.99$

The standard deviation of the data

$6,5,9, 13, 12, 8, 10$ is

0%
$\dfrac{\sqrt{52}}{7}$
0%
$\dfrac{52}{7}$
0%
$\dfrac{\sqrt{53}}{7}$
0%
$\dfrac{53}{7}$
0%
$6$

Explanation

Given data

$6,5,9,13,12,8,10$ Mean of the given data

$(\bar{x})$

$=\dfrac{6+5+9+13+12+8+10}{7}$

$=\dfrac{63}{7}=9$

The deviation of the respective data from the mean i.e.

$(x_i-\bar{x})$ are

$6-9,5-9,9-9,13-9,12-9,8-9,10-9$

$(x_i-\bar{x})=-3,-4,0,4,3,-1,1$

$(x_1-\bar{x})^2=9,16,0,16,9,1,1$

$\displaystyle \sum^7_{i=i} (x_1-\bar{x})^2=9+16+0+16+9+1+1=52$

$\therefore$ standard deviation

$(\sigma)$

$\displaystyle =\sqrt{\dfrac{1}{n}\sum^7_{i=1} (x_1-\bar{x})^2}=\sqrt{\dfrac{52}{7}}$

The mean and the standard deviation (s.d) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by

$p$ and then reduced by

$q$ , where

$p \neq 0$ and

$q \neq 0$ . If the new mean and new s.d. become half of their original values, then

$q$ is equal to:

0%
$-10$
0%
$-5$
0%
$-20$
0%
$10$

Explanation

${\textbf{Step -1: Let initial mean}}$ ${\mathbf{\left( {\overline x } \right)}}$ ${\textbf{and standard deviation}}$ ${\mathbf{\left( {{\sigma _1}} \right)}}$ ${\textbf{of 10 observation are 20 and 2 respectively.}}$

${\text{Now, each of these observations is multiplied by p and reduced by q.}}$

${\text{Thus, The new mean}}$ $= \overline {{x_1}} = p\overline x - q \ldots \left( 1 \right)$

${\text{Also, it is given that the new mean is half of the original mean}}{\text{.}}$

$\Rightarrow \overline {{x_1}} = \dfrac{1}{2}\overline x = \dfrac{1}{2} \times 20$

$\Rightarrow \overline {{x_1}} = 10$

${\text{Substitute this value of new mean in equation 1.}}$

$\Rightarrow 10 = p\left( {20} \right) - q$

$\Rightarrow 20p - q = 10 \ldots \left( 2 \right)$

${\textbf{Step -2: Find the value of p and q using the standard deviation.}}$

${\text{New standard deviation is given by,}}$

${\sigma _2} = \left| p \right|{\sigma _1} \ldots \left( 3 \right)$

${\text{As it will not be affected by subtraction of q from each observation.}}$

${\text{It is given that new standard deviation is half of the original.}}$

$\Rightarrow {\sigma _2} = \dfrac{1}{2}{\sigma _1} = \dfrac{1}{2} \times 2 = 1$

${\text{Substitute this value in equation 3.}}$

$\Rightarrow \left| p \right| \times 2 = 1$

$\Rightarrow p = \pm \dfrac{1}{2}$

${\textbf{Step -3: Find the value of q using p.}}$

${\text{If }}$ $p = \dfrac{1}{2}$

${\text{Then from equation 2 we have,}}$

$\Rightarrow 20 \times \dfrac{1}{2} - q = 10$

$\Rightarrow 10 - 10 = q$

$\Rightarrow q = 0$ $[\textbf{Rejected, as q}\neq 0]$

${\text{If }}$ $p = - \dfrac{1}{2}$

${\text{Using equation 2, we have,}}$

$\Rightarrow 20 \times \left( { - \dfrac{1}{2}} \right) - q = 10$

$\Rightarrow q = - 10 - 10$

$\Rightarrow q = - 20$

${\textbf{Hence, option C. i.e.}}{\mathbf{\left ( -20 \right )}} {\textbf{ is the correct answer.}}$

In a bar graph, each bar (rectangle) represents only one value of the numerical data.

0%
True
0%
False

In a bar graph, bars of uniform width are drawn
vertically only.

0%
True
0%
False

In a bar graph, the gap between two consecutive bars may not be the same.

0%
True
0%
False

The following bar graph represents the data for different sizes of shoes worn by the students in a school. Read the graph and state whether true or false:
The total number of students wearing shoe sizes

$5$ and

$8$ is
the same as the number of students wearing shoe size

$6$ .

0%
True
0%
False

Explanation

The given bar graph represents different shoe size worn by the students in a school.

We can prepare frequency table from given bar graph as follows:

Shoe size No. of students

$4$ $250$

$5$ $200$

$6$ $300$

$7$ $400$

$8$ $150$

Total number of students $=1300$ .

Here, the total number of students wearing shoe size $5$ and $8$

$=$ Number of students wearing shoe size $5$ $+$ Number of students wearing shoe size $8$

$=200+150=350$ students.

Also, Number of students wearing shoe size $6=300$ students.

Since, $350\ne300$ ,

we can say that the total number of students wearing shoe size $5$ and $8$ is not same as the number of students wearing shoe size $6$ .

That is the given statement is false and option $B$ is correct.

Data was collected on a student's typing rate and graph was drawn as shown below. Approximately how many words had this student typed in

$30$ seconds?

0%
$20$
0%
$24$
0%
$28$
0%
$34$

Explanation

The correct answer is option (c). approximately

$28$ words were typed in

$30$ seconds.

Observe the adjoining bar graph, showing the number of one-day international matches played by cricket teams of different countries. Choose the correct answer from the given four options:
How many matches did South Africa play?

0%
$16$
0%
$18$
0%
$20$
0%
$24$

Explanation

The given bar graph represents the number of matches played by cricket teams of different country.

We can prepare frequency table from given bar graph as follows:

Country No. of matches

India $30$

Pakistan $24$

West Indies $20$

England $28$

South Africa $18$

Australia $32$

Sri Lanka $24$

Clearly, the number of matches played by South Africa is $18$ .

Hence, option $B$ is correct.

0%
$6$
0%
$12$
0%
$24$
0%
$30$

Explanation

The given bar graph represents the number of matches played by cricket teams of different country.

We can prepare frequency table from given bar graph as follows:

Country No. of matches

India $30$

Pakistan $24$

West Indies $20$

England $28$

South Africa $18$

Australia $32$

Sri Lanka $24$

Total number of students $=176$ .

Then, the number of more matches played by India than Pakistan

$=$ Number of matches played by India $-$ Number of matches played by Pakistan

$=30-24=6$ matches.

That is, India played $6$ matches more than Pakistan.

Hence, option $A$ is correct.

Observe the following pictograph which shows the number of ice cream cones sold by school canteen during a week. Chose the correct answer from the given options:
Ratio of the number of ice cream cones sold on Saturday to the number of ice cream cones sold on Wednesday is.

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

Explanation

Saturday =

$4 \times 2= 8$
Wednesday =

$6 \times 2 = 12$

$\therefore$ Required ratio is

$8 : 12$

$\Rightarrow 2 : 3$

0%
India
0%
England
0%
Pakistan
0%
Australia

Explanation

The given bar graph represents the number of matches played by cricket teams of different country.

We can prepare frequency table from given bar graph as follows:

Country No. of matches

India $30$

Pakistan $24$

West Indies $20$

England $28$

South Africa $18$

Australia $32$

Sri Lanka $24$

Total number of students $=176$ .

The country that played maximum number of matches will have the highest number of matches.

Clearly, the maximum number of matches are played by Australia $(32)$ .

Hence, option $D$ is correct.

Observe the following pictograph which shows the number of ice cream cones sold by school canteen during a week. Chose the correct answer from the given options:
Total number of ice cream cones sold during the whole week was:

0%
$33$
0%
$67$
0%
$65$
0%
$57$

Explanation

From given pictograph, we can find the number of ice cream cons sold during a week as following

Monday = 10
Tuesday = 16
Wednesday = 12
Thursday = 7
Friday = 14
Saturday = 8

$\therefore$ Total number of ice cream

$= 67$ .

Observe the following pictograph which shows the number of ice cream cones sold by school canteen during a week. Chose the correct answer from the given options:
If the cost of one ice cream cone is

$Rs. 20,$ then the sale value on friday was:

0%
Rs. $70$
0%
Rs. $140$
0%
Rs. $280$
0%
Rs. $1340$

Explanation

Number of ice creams sold on friday

$= 7 \times 2$

$= 14$

$\therefore$ Sale value on Thursday

$= Rs.20 \times 14$

$= Rs. 280$

0%
$4 : 5$
0%
$5 : 4$
0%
$4 : 3$
0%
$7 : 6$

Explanation

The given bar graph represents the number of matches played by cricket teams of different country.

We can prepare frequency table from given bar graph as follows:

Country No. of matches

India $30$

Pakistan $24$

West Indies $20$

England $28$

South Africa $18$

Australia $32$

Sri Lanka $24$

Total number of students $=176$ .

Then, the ratio of number of matches played by India to the number of matches played by Sri Lanka

$=\dfrac{\text{Number of matches played by India}}{\text{Number of matches played by Sri Lanka}}$

$=\dfrac{30}{24}=\dfrac{5}{4}=5:4$ .

Hence, the required ratio is $5:4$ .

Therefore, option $B$ is correct.

In a series

$\sum x^{2} = 100, n = 5$ and

$\sum x = 20$ , then standard deviation is

0%
16
0%
2
0%
4
0%
8

Explanation

Standard deviation

$\sigma \sqrt{\dfrac{\sum x^{2}}{n} - \left ( \dfrac{\sum }{n} \right )^{2}} = \sqrt{\dfrac{100}{5} - \left ( \dfrac{20}{5} \right )^{2}}$

$\sqrt{20-16} = \sqrt{4} = 2$

Standard deviation of data

$6, 10, 4, 7, 4, 5$ is -

0%
$\sqrt{\dfrac{13}{3}}$
0%
$\dfrac{13}{3}$
0%
$\sqrt{26}$
0%
$\dfrac{\sqrt{26}}{6}$

Explanation

Mean

$(\bar{x}) = \dfrac{6+ 10 + 4 + 7 + 4 + 5}{6} = \dfrac{36}{6} = 6$
Standard deviation

$(\sigma ) = \sqrt{\dfrac{1}{N} \times \sum (x_{1} - \bar{x})^{2}}$

$=\sqrt{\dfrac{1}{6} \times 26} = \sqrt{\dfrac{13}{3}}$

If the student wants to raise his total score upto

$225$ , then how many more marks should he score?

0%
$125$
0%
$100$
0%
$200$
0%
$25$

The exam scores of all

$500$ students were recorded and it was determined that these scores were normally distributed. If Jane's score is

$0.8$ standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?(Area under the curve below

$z=0.8 \ is \ 0.7881$ )

0%
$394$
0%
$250$
0%
$400$
0%
$106$

Explanation

Let the Jane's score be

$x$ .

Let

$m$ be the mean and

$s$ be the standard deviation and then we find the

$z$ score.

Since it is given that Jane's score is

$0.8$ standard deviation above the mean means

$x=0.8s+m$

$z=\dfrac { x-m }{ s } =\dfrac { 0.8s+m-m }{ s } =0.8\\$

$\Rightarrow z=0.8$

The percentage of students who scored above Jane is (from table of normal distribution) is

$1-0.7881=0.2119=21.19$

$\%$

The number of students who scored above Jane is (from table of normal distribution) is

$\dfrac { 21.9 }{ 100 } \times 500=106$

Hence,

$106$ students scored above Jane.

Find the variance of the series 5, 8, 11, 14 and 17.....

0%
17
0%
18
0%
16
0%
12

Explanation

Given series is

$5, 8, 11, 14, 17$
Mean

$=\cfrac { 5+8+11+14+17 }{ 5 } =\cfrac { 55 }{ 5 }$
Mean

$=\mu=11$
Variance

$=\sum { { { \left( { x }_{ i }-\mu \right) }^{ 2 } }/{ n } }$

$=\cfrac { 1 }{ 5 } \left[ \left( 5-11 \right) ^{ 2 }+{ \left( 8-11 \right) }^{ 2 }+{ \left( 11-11 \right) }^{ 2 }+{ \left( 14-11 \right) }^{ 2 }+{ \left( 17-11 \right) }^{ 2 } \right]$

$=\cfrac { 1 }{ 5 } \left[ 36+9+0+9+36 \right]$

$=\cfrac { 90 }{ 5 }$

$=18$

The probability distribution of a random variable

$X$ is given below:

$X=x$	0	1	2	3
$P(X=x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

Then the variance of

$X$ is

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

Explanation

$E[x^2]=0+1^2\cdot \displaystyle \frac{2}{10}+2^2\cdot \frac{3}{10}+3^2\cdot \frac{4}{10}=\frac{[2+12+36]}{10}=5.0$

$E[x]=0+\displaystyle \frac{2}{10}+2\cdot \frac{3}{10}+3\cdot \frac{4}{10}=\frac{[2+6+12]}{10}=2$

$\therefore$ Var (x)

$= E[x^2] - E[x]^2 = 5-2^2=1$

Which two years did the least number of boys attend the convention?

0%
1995 and 1996
0%
1995 and 1998
0%
1996 and 1997
0%
1996 and 1992
0%
1997 and 1998

The federal government in the United States has the authority to protect species whose populations have reached dangerously low levels. The figure above represents the expected populations of a certain endangered species before and after a proposed law aimed at protecting the animal is passes. Based on the graph, which of the following statements is true?

0%
The proposed law is expected to accelerate the decline in population.
0%
The proposed law is expected to stop and reverse the decline in population.
0%
The proposed law is expected to have no effect on the decline in population.
0%
The proposed law is expected to slow, but not stop or reverse, the decline in population.

Explanation

The federal government in the United States has the authority to protect species whose populations have reached dangerously low levels. Based on the given graph, we can say that the proposed law is expected to slow, but not stop or reverse the decline in population.

Hence, option

$D$ is correct.

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 the combined data set is

0%
$\displaystyle \frac{11}{2}$
0%
$6$
0%
$\displaystyle \frac{13}{2}$
0%
$\displaystyle \frac{5}{2}$

Explanation

$\sigma_{\mathrm{x}}^{2}=4$

$\sigma_{\mathrm{y}}^{2}=5$

$\overline{\mathrm{x}}=2$

$\overline{\mathrm{y}}=4$

$\displaystyle \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{5}=2, \Sigma \mathrm{x}_{\mathrm{i}}=10;\Sigma \mathrm{y}_{\mathrm{i}}=20$

$\sigma_{\mathrm{x}}^{2}=(\frac{1}{5}\Sigma \mathrm{x}_{\mathrm{i}}^{2})-(\overline{x})^{2}=\frac{1}{5}(\Sigma \mathrm{x}_{1}^{2})-4$

$\sigma_{\mathrm{y}}^{2}=(\frac{1}{5}\Sigma \mathrm{y}_{\mathrm{i}}^{2})-(\overline{y})^{2}=\frac{1}{5}(\Sigma \mathrm{y}_{1}^{2})-4$

$\Sigma \mathrm{x}_{\mathrm{i}}^{2}=40$

$\Sigma \mathrm{y}_{\mathrm{i}}^{2}=105$

$\sigma_{\mathrm{z}}^{2}=\displaystyle \frac{1}{10}(\Sigma \mathrm{x}_{\mathrm{i}}^{2}+\Sigma \mathrm{y}_{\mathrm{i}}^{2})-(\frac{\overline{\mathrm{x}}+\overline{\mathrm{y}}}{2})=\frac{1}{10}(40+105)-9=\frac{145-90}{10}=\frac{55}{10}=\frac{11}{2}$

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 10 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 10 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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