MCQ Questions for Class 9 Maths Statistics Quiz 4

The mean of

$a, b, c, d$ and

$e$ is

$28$ . If the mean of

$a, c$ and

$e$ is

$24$ , what is the mean of

$b$ and

$d$ ?

0%
$31$
0%
$32$
0%
$33$
0%
$34$

Explanation

Formula used:

$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$

Given

$\Rightarrow$

$\dfrac{a+b+c+d+e}{5}=28$ and

$\dfrac{a+c+e}{3} =24$

To find

$\Rightarrow$

$\dfrac{b+d}{2}$ which is nothing but the mean of

$b$ and

$d$ .

$\begin{aligned}{}\dfrac{{a + b + c + d + e}}{5} &= 28\\a + b + c + d + e& = 140\quad \dots (i)\end{aligned}$

Also,

$\begin{aligned}{}\dfrac{{a + c + e}}{3} &= 24\\a + c + e& = 72\quad \dots (ii)\end{aligned}$

By using

$(i)$ and

$(ii)$ ,

$\begin{aligned}{}a + b + c + d + e &= 140\\b + d + 72 &= 140\\b + d &= 68\\\dfrac{{b + d}}{2} &= 34\end{aligned}$

Hence, option

$D$ is the correct answer.

For which state the average number of candidates selected over the years is the maximum?

0%
Delhi
0%
H.P
0%
U.P
0%
Punjab

Explanation

The average number of candidates selected over the given period for various states are:

For Delhi

$= \dfrac {94 + 48 + 82 + 90 + 70}{5}$

$= \dfrac {385}{5} = 76.8$

For U.P.

$= \dfrac {78 + 85 + 48 + 70 + 80}{5}$

$= \dfrac {361}{5} = 72.2$

For Punjab

$= \dfrac {85 + 70 + 65 + 84 + 60}{5}$

$= \dfrac {364}{5} = 72.8$

For Haryana

$= \dfrac {75 + 75 + 55 + 60 + 75}{5}$

$= \dfrac {340}{5} = 68$

The average is maximum for Delhi.

The librarian at the public library counted the number of books on each shelf. The highest number of books contained by any of the self is:

0%
$0$
0%
$2$
0%
$6$
0%
$8$

Explanation

On looking the graph, we can say that the books are high at the intervals

$11-20$ and

$71-80$ as

$6$ (The highest bar has the highest frequency).

Hence, option

$C$ is correct.

What is the average amount of interest per year which the company had to pay during this period?

0%
$Rs. 32.43$ lakhs
0%
$Rs. 33.72$ lakhs
0%
$Rs. 34.18$ lakhs
0%
$Rs. 36.66$ lakhs

Explanation

Average amount of interest paid by the Company during the given period

$= Rs. \left [\dfrac {23.4 + 32.5 + 41.6 + 36.4 + 49.4}{5}\right ]lakhs$

$= Rs. \left [\dfrac {183.3}{5}\right ]lakhs$

$= Rs. 36.66\ lakhs$

Mean of marks obtained by

$10$ students is

$30$ .
Marks obtained are

$25,30,21,55,47,10,15,x,45,35$ .
Find the value of

$x$ .

0%
$25$
0%
$37$
0%
$69$
0%
$17$

Explanation

Let

$u$ be the average marks of the class.

$\therefore u=\dfrac{25+30+21+55+47+10+15+x+45+35}{10}=\dfrac{283+x}{10}$
But

$u=30$ .
So,

$30=\dfrac{283+x}{10}$ .

$\therefore x=17$
Ans-Option D

Find AM of divisors of

$100$ .

0%
$24$
0%
$25.5$
0%
$24.11$
0%
$21.9$

Explanation

Divisors of 100 are:

$1, 2, 4, 5, 10, 20, 25, 50, 100$

$\therefore n=9$

$\therefore S=1+2+4+5+10+20+25+50+100=217$

$\therefore A.M.=\dfrac{S}{n}=\dfrac{217}{9}=24.11$

If the Arithmetic mean of

$8, 6, 4, x, 3, 6, 0$ is

$4$ ; then the value of

$x =$

0%
$7$
0%
$6$
0%
$1$
0%
$4$

Explanation

Arithmetic mean

$= \cfrac{\text{sum of all observations}}{\text{no. of observations}}$

$\Rightarrow 4=\cfrac { 8+6+4+x+3+6+0 }{ 7 } \\ \Rightarrow 28=27+x\\ \Rightarrow x=1$

The relative frequency of students working

$9$ hours or less is

0%
$2$
0%
$40$
0%
$45$
0%
Can not be determined from the information given.

Explanation

The relative frequency of students working

$9$ hours or less lie in the

$0-9$ hours group.
So, total number of students

$=40$ .

The histogram shows the birth weights of 100 new born babies.
How many babies weighed 8 lb or more?

0%
$23$
0%
$45$
0%
$30$
0%
$22$

Explanation

Given graph shows birth weights of 100 new born babies

From the graph, the number of babies whose weight is

$8\ lb$ or more is

$22+16+6+1=45$

Find the number of students working

$19$ hours or less.

0%
$40$
0%
$50$
0%
$90$
0%
Can not be determined without the original data

Explanation

The number of students working

$19$ hours or less will contain the group which works for

$0-9$ hours as well as

$10-19$ hours.
So, total students

$=40+50=90$ .

The mean age of

$25$ students in a class is

$12$ . If the teacher's age is included, the mean age increases by

$1$ . Then teacher's age (in years) is:

0%
$24$
0%
$38$
0%
$30$
0%
$48$

Explanation

Sum of

$25$ students ages be

$x$

$\Rightarrow$ Mean

$=\dfrac {x}{25}=12\Rightarrow x=300$ equation - (1)

Let the teacher's age be

$y$

$\Rightarrow$ New mean

$=\dfrac {x+y} {26}=13$

$\Rightarrow 300+y=338$

$\Rightarrow y=38$

Teacher's age

$=y=38$

The mean salary paid per week to

$1000$ employees of an establishment was found to be Rs.

$900$ . Later on, it was discovered that the salaries of two employees were wrongly recorded as Rs.

$750$ and Rs.

$365$ instead of Rs.

$570$ and Rs.

$635$ . Find the corrected mean salary.

0%
$900.90$
0%
$1,115$
0%
$1,225$
0%
$900.09$

Explanation

We first find the corrected sum of observation.
Corrected sum of observations

$=$ (Sum of total incorrect observation)

$-$ (Sum of incorrect data)

$+$ (Sum of correct data).

$=900\times 1,000-(750+365)+(570+635)=9,00,090$

$\therefore$ Corrected Mean

$=Rs.\dfrac{9,00,090}{1,000}=Rs.900.09$

Find the arithmetic mean of numbers from

$1$ to

$9$ .

0%
$9$
0%
$5$
0%
$8$
0%
$3$

Explanation

The numbers from

$1$ to

$9$ are

$1,2,3,4,5,6,7,8,9$

$A.M. =\dfrac{\text{Sum of observations}}{\text{Number of observations}}= \dfrac {1+2+3+4+5+6+7+8+9}{9} = \dfrac {45}{9} = 5$

The mean of

$\displaystyle\frac{1}{3},\frac{3}{4},\frac{5}{6},\frac{1}{2}$ and

$\displaystyle\frac{7}{12}$ is ____________.

0%
$\displaystyle\frac{2}{5}$
0%
$\displaystyle\frac{3}{5}$
0%
$\displaystyle\frac{1}{5}$
0%
None of these

Explanation

Given data is

$\dfrac{1}{3},\dfrac{3}{4},\dfrac{5}{6},\dfrac{1}{2},\dfrac{7}{12}$

Mean

$=$ Sum of observations

$\div$ Total number of observations

Sum of observations

$=$

$\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{5}{6}+\dfrac{1}{2}+\dfrac{7}{12}$

Mean

$=$

$\dfrac{\dfrac{4+9+10+6+7}{12}}{5}$

$= \dfrac{3}{5}$

Mean of

$9$ observations was founded to be

$35$ . Later on, it was detected that an observation

$81$ was misread as

$18$ , then the correct mean of the observations is ____________.

0%
$40$
0%
$41$
0%
$42$
0%
$43$

Explanation

Mean of

$9$ observations

$= 35$

Sum of

$9$ observations

$= 35\times9 = 315$

Since

$81$ was misread as

$18$

New sum

$= 315-18+81= 378$

The correct mean

$= \dfrac{378}{9} = 42$

Primary data in statistics is the data collected from :

0%
Primary sources
0%
Secondary sources
0%
Both
0%
None of the above

The mean weight of 9 students is 25 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is

0%
25 kg
0%
24 kg
0%
26 kg
0%
23 kg

Explanation

Formula used:

$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$

Apply the above formula, we get the sum of weights of the 9 students =

$25\times 9=225$ kg

If one more student is joined in the group, then total number of students is 10, and the mean isn 25.

The sum of weights of the 10 students is

$25\times 10=250$ kg

The weight of the tenth student is

$250-225=25$ kg

The sum of

$15$ observations is

$434+x.$ If the mean of the data is

$x,$ then find

$x.$

0%
$25$
0%
$27$
0%
$31$
0%
$33$

Explanation

Given, sum of all observations

$=434+x$

Number of observations

$=15$

and

$Mean=x$

$\Rightarrow \dfrac{434+x}{15}=x$

$\Rightarrow 434+x=15x$

$\Rightarrow x-15x=-434$

$\Rightarrow -14x=-434$

$\Rightarrow x=\dfrac{-434}{-14}$

$\Rightarrow x=31$

Hence, the value of

$x$ is

$31$ .

In a frequency distribution with classes

$(1-8),(9-16),(17-24)....$ the class interval is

0%
8
0%
9
0%
7
0%
3

Explanation

Class interval

$=$ (high

$-$ low value

$+1$ )

$=8-1+1$

$=8$ .

1127948_1242003_ans_40efca94ead14fc1b6da0ab5432908a9.jpg

Range of the data is always from the data.

0%
True
0%
False

From the given histogram, we can say that

$1500$ males above the age of

$20$ are literate.

0%
True
0%
False

Explanation

The given figure represents the graph showing the number of books read by number of people.

The given statement says about the number of males who are literate based on the age of the person.

Hence, the given statement is false.

Therefore, option

$B$ is correct.

State whether the following statements are true (T) or false (F):
In a bar graph, width of a bar has no significance. It is only for eye attraction.

0%
True
0%
False

Explanation

We know,

A bar graph is a graphical representation of data using bars of equal width drawn vertically or horizontally with equal spacing between them and different heights.

But, width of a bar do not have any significance and is only for eye attraction.

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the following statements are true (T) or false (F):
In a bar graph, bars can be drawn either vertically or horizontally.

0%
True
0%
False

Explanation

We know,

A bar graph is a graphical representation of data using bars of equal width drawn vertically or horizontally with equal spacing between them and different heights.

Then clearly, in a bar graph, bars can be drawn either vertically or horizontally.

That is, the statement is true and option

$A$ is correct.

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:

0%
$2m + l$
0%
$2m - l$
0%
$m + l$
0%
$m + 2l$

Explanation

Let

$a$ be lower class limit,

$l$ is the upper class limit and

$m$ is the mid-point
then, mid-point

$m = \dfrac{a +l}{2}$
Therefore,

$a = 2m - l$

If the fifth class interval is

$60 -65,$ fourth class interval is

$55-60,$ then the first class interval is

$45-50$ .

0%
True
0%
False

Explanation

The fifth interval is

$60 -65$ and fourth class interval is

$55-60 ,$ then third class interval is

$50-55 ,$ second class interval is

$45-50$ and first class interval is

$40-45$ . Hence, the given statement is false.

Which of the following data is not primary?

0%
By visiting a certain class, gathering information about the attendance of students.
0%
By an actual visit to homes, to find a number of family members.
0%
To get information regarding the plantation of soyabean done by each farmer from the village Talathi.
0%
Review the cleanliness status of canals by actually visiting them.

Fill in the blank:
The height of a rectangle in a histogram represent _______.

0%
class interval
0%
midpoint
0%
frequency density
0%
frequency

Explanation

We know, histogram is a graphical representation of data using bars of equal width and no gaps between them. The height of the bars gives the frequency of the particular class interval.

Therefore, the height of a rectangle in a histogram represents the frequency.

Hence, option

$D$ is correct.

The class mark of the class interval 20 - 29 is

0%
20
0%
29
0%
25
0%
24.5

Explanation

The class mark of the class interval 20 - 29 is

$\dfrac{20+29}{2}=24.5$

The mean of first ten odd natural number is

0%
5
0%
10
0%
20
0%
19

Explanation

Since, first ten odd natural numbers are $1, 3, 5, 7, 9, 11, 13, 15, 17, 19$

We know that

$\text{Mean}=\dfrac{\text{sum of observation}}{\text{number of observation}}$

$\therefore$ Mean $=\dfrac{(1 + 19 + 3 + 17 + 5 + 15 + 7 + 13 + 9 + 11) }{ 10} = \dfrac{100}{10} = 10$

What is the upper-class limit for the class

$25-35$ ?

0%
$25$
0%
$35$
0%
$60$
0%
$30$

CBSE Questions for Class 9 Maths Statistics Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Statistics Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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