Statistics Test

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Statistics Test - 38

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Weekly Quiz Competition

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 $$\displaystyle\frac{N}{2}(40.5)$$ is $$61$$ and value of $$x$$ corresponding to $$61$$ is $$5$$.

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

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SHARING IS CARING

Question 1

$$9$$

$$11$$

$$12$$

$$15$$

Solution

Question 2

$$5$$

$$6$$

$$4$$

$$7$$

Solution

Question 3

38

42

67

51

Solution

Question 4

$$55$$

$$65$$

$$45$$

$$95$$

Solution

Question 5

$$3$$

$$2$$

$$1$$

$$7$$

Solution

Question 6

$$5$$

$$7$$

$$6$$

$$12$$

Solution

Question 7

$$18.9$$

$$17.9$$

$$16.94$$

$$14.5$$

Solution

Question 8

$$75$$

$$30$$

$$60$$

$$25$$

Solution

Question 9

$$7$$

$$7.5$$

$$8$$

$$8.5$$

Solution

Question 10

Rs. 2.12

Rs. 2.06

Rs. 2.20

Rs. 2.22

Solution

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