Statistics Test

Result

Statistics Test - 38

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Question 1
1 / -0

Calculate the mode
$x$ 3 6 9 12 15 18
$f$ 6 8 11 4 10 7

A
$9$

B
$11$

C
$12$

D
$15$

Solution

The variate $(x)$ is with highest frequency $11$ .
So, mode = $9$

Question 2

1 / -0

Find the median of the following distribution

x	1	2	3	4	5	6	7
f	7	13	12	8	21	9	11

$5$

$6$

$4$

$7$

Solution

$x$	$f$	Cumulative Frequency
1	$7$	$7$
2	$13$	$7+13 = 20$
3	$12$	$20+12 = 32$
4	$8$	$32+8 = 40$
5	$21$	$40+21 = 61$
6	$9$	$61+9 = 70$
7	$11$	$70+10 = 81$
Here $N = 81$		$\Rightarrow \displaystyle\frac{N}{2} = 40.5$

We find that the cummulative frequency greater than

\displaystyle\frac{N}{2}(40.5)

61

and value of

x

corresponding to

61

5

.

Hence,

Median = 5

Question 3
1 / -0

The contents of 100 match boxes were checked to determine the number of matches they contained.
No. of matches 35 36 37 38 39 40 41
No. of match boxes 6 10 12 15 11 7 8

Calculate the mean number of matches per box

A
38

B
42

C
67

D
51

Solution

In order to determine the mean, we use the formula as below

$Mean = \displaystyle\frac{x_1f_1+ x_2f_2 + x_3f_3 + x_4f_4 + x_5f_5 + x_6f_6 + x_7f_7}{f_1+ f_2 + f_3 + f_4 + f_5 + f_6 + f_7}$

$\quad = \displaystyle\frac{(35\times6) + (36\times10) + (37\times12) + (38\times15) + (39\times11) + (40\times7) + (41\times8)}{6+10+12+15+11+7+8}$
$= \displaystyle\frac{210+360+444+570+429+280+328}{69} = 38$
Question 4
1 / -0

Find the mean
x: 10 30 50 70 89
f: 7 8 10 15 10

A
$55$

B
$65$

C
$45$

D
$95$

Solution

$Mean = \displaystyle\frac{x_1f_1+ x_2f_2 + x_3f_3 + x_4f_4 + x_5f_5}{f_1+ f_2 + f_3 + f_4 + f_5}$

$\quad = \displaystyle\frac{10\times7+30\times8+50\times10+70\times15+89\times10}{7+8+10+15+10} = \displaystyle\frac{2750}{50} = 55$
Question 5
1 / -0

Find the value of $x$ , if the mean of the following distribution is $7.5$
x: 3 5 7 9 11 13
f: 6 8 15 x 8 4

A
$3$

B
$2$

C
$1$

D
$7$

Solution

$Mean = \displaystyle\frac{x_1f_1+ x_2f_2 + x_3f_3 + x_4f_4 + x_5f_5 + x_6f_6}{f_1+ f_2 + f_3 + f_4 + f_5 + f_6}$

$\quad 7.5 = \displaystyle\frac{18+40+105+9x+88+52}{41+x}$

$\Rightarrow \quad 7.5(41+x) = 303 + 9x \\\Rightarrow 307.5 +7.5x=303+9x\\ \Rightarrow x = 3$
Question 6
1 / -0

If the mean of the following distribution is 6, then value of p is
$x: 2\ 4\ 6\ 10\ p+5$
$f: 3\ 2\ 3\ 1\ ~~2$

A
$5$

B
$7$

C
$6$

D
$12$

Solution

$6=\dfrac {(2\times 3)+(4\times 2)+(6\times 3)+(10\times 1)+(p+5)\times 2}{3+2+3+1+2}$

$\Rightarrow 6=\dfrac{6+8+18+10+2p+10}{11}$

$\Rightarrow 6=\dfrac{52+2p}{11}$
$\Rightarrow 6\times11=52+2p$
$\Rightarrow 66-52=2p$
$\Rightarrow 14=2p$

$\Rightarrow p=\dfrac{14}{2}$
$\therefore p=7$
Question 7
1 / -0

If median = $20.6$ , mode = $26$ then find mean

A
$18.9$

B
$17.9$

C
$16.94$

D
$14.5$

Solution

We know that
Mode = 3 Median - 2 Mean
So Mean = $\displaystyle \frac{1}{2}\left [ 3Median-Mode \right ]$
= $\displaystyle \frac{1}{2}\left [ 3\times 20.6-26 \right ]$
= $\displaystyle \frac{1}{2}\left [ 35.8 \right ]$
= 17.9
Question 8
1 / -0

In a class test in English $10$ students scored $75$ marks $12$ students scored $60$ marks $8$ scored $40$ marks and $3$ scored $30$ marks the mode for their score is:

A
$75$

B
$30$

C
$60$

D
$25$

Solution

$\textbf{Step -1: Defining mode.}$
$\text{In a given data set, the mode is the value that appears most often among every other values.}$
$\text{Given data set says, }$
$\text{10 Students scored 75 marks.}$
$\text{12 Students scored 60 marks.}$
$\text{8 Students scored 40 marks.}$
$\text{3 Students scored 30 marks.}$
$\textbf{Step -2: Finding the mode of their score.}$
$\text{The mode of their score will be the marks occurring most often.}$
$\therefore \text{12 Students scored 60 marks.}$
$\textbf{Hence, The mode of their score is 60. (Option C)}$
Question 9
1 / -0

x 3 5 7 12 8 9
f 7 10 13 10 5 12
The median of the above data is

A
$7$

B
$7.5$

C
$8$

D
$8.5$

Solution

Median $\displaystyle =\left ( \dfrac{57+1}{2} \right )^{th}value=29th =7$
Or
We find that the cummulative frequency greater than $\displaystyle\frac{N}{2}(28.5)$ is $30$ and value of $x$ corresponding to $30$ is $7$ .

Hence, $Median = 7$
Question 10
1 / -0

Factory A Factory B
No. of wage earners 250 200
Average daily wage Rs. 2.00 Rs. 2.50
The average of daily wages for the earners of the two factories combined is

A
Rs. 2.12

B
Rs. 2.06

C
Rs. 2.20

D
Rs. 2.22

Solution

Required average
$=\displaystyle \frac{250 \times 2.00 + 200 \times 2.50}{250 + 200} = \frac{1000}{450} =\frac{20}{9} =$ Rs. 2.22

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Re-Attempt Weekly Quiz Competition

	Factory A	Factory B
No. of wage earners	250	200
Average daily wage	Rs. 2.00	Rs. 2.50

No. of matches	35	36	37	38	39	40	41
No. of match boxes	6	10	12	15	11	7	8

Statistics Test - 38

Result

Statistics Test - 38

Score

-

Rank

-

SHARING IS CARING

Question 1

999

111111

121212

151515

Solution

Question 2

555

666

444

777

Solution

Question 3

38

42

67

51

Solution

Question 4

555555

656565

454545

959595

Solution

Question 5

333

222

111

777

Solution

Question 6

555

777

666

121212

Solution

Question 7

18.918.918.9

17.917.917.9

16.9416.9416.94

14.514.514.5

Solution

Question 8

757575

303030

606060

252525

Solution

Question 9

777

7.57.57.5

888

8.58.58.5

Solution

Question 10

Rs. 2.12

Rs. 2.06

Rs. 2.20

Rs. 2.22

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

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$9$

$11$

$12$

$15$

$5$

$6$

$4$

$7$

$55$

$65$

$45$

$95$

$3$

$2$

$1$

$7$

$5$

$7$

$6$

$12$

$18.9$

$17.9$

$16.94$

$14.5$

$75$

$30$

$60$

$25$

$7$

$7.5$

$8$

$8.5$