Statistics Test

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Statistics Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Find median of the data, using empirical relation when it is given that mode $$=12.4$$ and mean $$=10.5$$.

A
$$11.10$$

B
$$11.13$$

C
$$12$$

D
$$13.20$$

Solution

Given, mode$$=12.4$$ and mean $$=10.5$$
Emperical formula is:
$$\text{mode} =3\times \text{median} - 2\times \text{mean}$$
Substituting the values, we get
$$12.4=3\times \text{median}$$ $$- 2(10.5)$$
$$\Rightarrow 12.4=3\times\text{ median}$$ $$- 21$$
$$\Rightarrow 3\times \text{median}$$ $$= 33.4$$
$$\Rightarrow \text{Median}$$ $$= 11.13$$
The median is $$ 11.13$$.
Question 2
1 / -0

If mean : median of a certain data is $$2 : 3$$, what is the ratio of its mode and mean?

A
$$3 : 2$$

B
$$5 : 2$$

C
$$3 : 5$$

D
$$2 : 3$$

Solution

$$\cfrac { mean }{ median } =\cfrac { 2 }{ 3 } \\ \cfrac { mode }{ mean } =?\\ \\ Since, \quad mode = \quad 3median - \quad 2mean \\ \quad mean-mode=3(mean-median)\\ \quad mean(1-\cfrac { mode }{ mean } )=3mean(1-\cfrac { median }{ mean } )\\ \Rightarrow 1-\cfrac { mode }{ mean } =3(1-\cfrac { 3 }{ 2 } )\\ \Rightarrow 1-\cfrac { mode }{ mean } =-\cfrac { 3 }{ 2 } \\ \Rightarrow 1+\cfrac { 3 }{ 2 } =\cfrac { mode }{ mean } $$

Therefore, $$\cfrac { mode }{ mean } =\cfrac { 5 }{ 2 } $$
Question 3
1 / -0

The demand for different shirt sizes is given in the table.
Size
38
39
40
41
42
43
44
No of Persons
26
36
20
15
13
7
5
Find the modal shirt size.

A
$$39$$

B
$$40$$

C
$$44$$

D
$$42$$

Solution

$$Size\\ \quad 38\\ \quad 39\\ \quad 40\\ \quad 41\\ \quad 42\\ \quad 43\\ \quad 44$$ $$No\quad of\quad person\\ \quad \quad \quad 26\\ \quad \quad \quad 36\\ \quad \quad \quad 20\\ \quad \quad \quad 15\\ \quad \quad \quad 13\\ \quad \quad \quad 07\\ \quad \quad \quad 05\\ $$

Therefore, number of person is maximum for size having $$39$$. Hence, modal size $$=39$$.
Question 4
1 / -0

If the ratio of mode and mean of a certain data is $$3 : 2$$, what is the ratio of its median and mean?

A
$$9 : 4$$

B
$$8: 9$$

C
$$7 : 6$$

D
$$4 : 9$$
Question 5
1 / -0

One of the methods to find out mode is ______.

A
Mode $$= 3$$ Median $$+ 2$$ Mean

B
Mode $$= 3 $$Median $$- 3$$ Mean

C
Mode $$= 2$$ Median $$- 3$$ Mean

D
Mode $$= 3$$ Median $$- 2$$ Mean

Solution

In order to find mode, we can use the relation, mode $$= 3$$ median $$- 2$$ mean.
Question 6
1 / -0

On the occasion of New year's day a sweet stall prepared sweet packets. Number of sweet packets and cost of each packet are given as follows.
Cost of packet (in Rs) $$Rs. 25$$ $$Rs. 50$$ $$Rs. 75$$ $$Rs. 100$$ $$Rs. 125$$ $$Rs. 150$$
No of packets $$20$$ $$26$$ $$32$$ $$29$$ $$22$$ $$11$$
Find the mean, median and mode of the data

A
$$\overline {x} = 82.1428; Median = 75; Mode = 75$$

B
$$\overline {x} = 70; Median = 75; Mode = 50$$

C
$$\overline {x} = 80; Median = 5; Mode = 50$$

D
None of these

Solution

Cost of packet Number of packets Cumulative frequency
25 20 20
50 26 46
75 32 78
100 29 107
125 22 129
150 11 140
To find the mean we will add up total cost of packers, then divide by total number of packets.
So, Mean $$= \dfrac{(25\times 20 + 50 \times 26 + 75\times 32+100\times 29+125\times 22+150\times 11)}{(20+26+32+29+22+11)}$$
Mean $$= \dfrac{(500 + 1300 +2400+ 2900+ 2750 + 1650)}{140} = \dfrac{11500}{140} = 82.1428$$
As we can see that the cost of greatest number of packets is Rs. 75 each, so, the mode will be 75.
As we can see there are 140 packets, arranged in increasing order of their cost. So, the middle one will be $$\dfrac{(140)}{2} = \dfrac{140}{2} = 70^{th}$$ packet in the list.
So, the median will be the cost of packet corresponding to cumulative frequency just greater or equal to $$70\ i.e. 75$$.
Therefore, median is $$75$$.
Question 7
1 / -0

For a certain frequency distribution, the value of Mean is $$101$$ and Median is $$100$$. Find the value of Mode.

A
$$98$$

B
$$102$$

C
$$100.5$$

D
$$99$$

Solution

We have

Mean $$= 101$$

Median $$= 100$$

We know that,

Mean $$-$$ Mode $$=$$ $$3($$Mean $$-$$ Median$$)$$

$$101 -$$ Mode $$= 3(101 - 100)$$

$$100 -$$ Mode $$= 3$$

$$101 - 3 =$$ Mode

Mode $$= 98$$
Question 8
1 / -0

For the next three (03) items that follow :
The number of telephone calls received in 245 successive one minute intervals at an exchange is given below in the following frequency distribution.
Number of calls 0 1 2 3 4 5 6 7
Frequency 14 21 25 43 51 40 39 12
What is the median of the distribution ?

A
$$3.5$$

B
$$4$$

C
$$4.5$$

D
$$5$$

Solution

Here the No.of calls are $$0, 1 , 2 , 3 , 4 , 5 , 6 , 7$$

First we should Place the data in ascending order (smallest to highest).
Here it is already in ascending order,

Now Add the two middle numbers and then divide by two, to get the average

$$\dfrac {(3+4)}{2} = 3.5$$
Hence The median is $$3.5.$$
Question 9
1 / -0

For the next three (03) items that follow :
The number of telephone calls received in 245 successive one minute intervals at an exchange is given below in the following frequency distribution.
Number of calls 0 1 2 3 4 5 6 7
Frequency 14 21 25 43 51 40 39 12
What is the mode of the distribution ?

A
3

B
4

C
5

D
6

Solution

Mode of a frequency distribution $$=$$ the value that appears most often in the set
From the given set, we can see that
$$4$$ occurs most ($$51$$ times)
$$\therefore mode =4$$
Question 10
1 / -0

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 15 13 17 10
For the information given above, $$\left( \dfrac{n}{2} - cf \right )$$ will be = ...............

A
$$10$$

B
$$20$$

C
$$30$$

D
$$25$$

Solution

Class Frequency Cummulative frequency (cf)
0-10 5 5
10-20 15 5 + 15 = 20
20-30 13 20+13 = 33
30-40 17 33 + 17 = 50
40-50 10 50 + 10 = 60
Hence $$ n = 60 $$
$$\Rightarrow \dfrac{n}{2} = 30$$
The cummulative frequency just greater than 30 is 33.
So, the class 20-30 will be the median class.
The cummulative frequency of the class preceding the median class is 20, that is cf = 20.
So, $$\dfrac{n}{2} - cf = 30 - 20 = 10$$

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Cost of packet (in Rs)	$$Rs. 25$$	$$Rs. 50$$	$$Rs. 75$$	$$Rs. 100$$	$$Rs. 125$$	$$Rs. 150$$
No of packets	$$20$$	$$26$$	$$32$$	$$29$$	$$22$$	$$11$$

Cost of packet	Number of packets	Cumulative frequency
25	20	20
50	26	46
75	32	78
100	29	107
125	22	129
150	11	140

Class	Frequency	Cummulative frequency (cf)
0-10	5	5
10-20	15	5 + 15 = 20
20-30	13	20+13 = 33
30-40	17	33 + 17 = 50
40-50	10	50 + 10 = 60

Size	38	39	40	41	42	43	44
No of Persons	26	36	20	15	13	7	5

Number of calls	0	1	2	3	4	5	6	7
Frequency	14	21	25	43	51	40	39	12

Number of calls	0	1	2	3	4	5	6	7
Frequency	14	21	25	43	51	40	39	12

Statistics Test - 51

Result

Statistics Test - 51

Score

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Rank

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SHARING IS CARING

Question 1

$$11.10$$

$$11.13$$

$$12$$

$$13.20$$

Solution

Question 2

$$3 : 2$$

$$5 : 2$$

$$3 : 5$$

$$2 : 3$$

Solution

Question 3

$$39$$

$$40$$

$$44$$

$$42$$

Solution

Question 4

$$9 : 4$$

$$8: 9$$

$$7 : 6$$

$$4 : 9$$

Question 5

Mode $$= 3$$ Median $$+ 2$$ Mean

Mode $$= 3 $$Median $$- 3$$ Mean

Mode $$= 2$$ Median $$- 3$$ Mean

Mode $$= 3$$ Median $$- 2$$ Mean

Solution

Question 6

$$\overline {x} = 82.1428; Median = 75; Mode = 75$$

$$\overline {x} = 70; Median = 75; Mode = 50$$

$$\overline {x} = 80; Median = 5; Mode = 50$$

None of these

Solution

Question 7

$$98$$

$$102$$

$$100.5$$

$$99$$

Solution

Question 8

$$3.5$$

$$4$$

$$4.5$$

$$5$$

Solution

Question 9

3

4

5

6

Solution

Question 10

$$10$$

$$20$$

$$30$$

$$25$$

Solution

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