Measures of Central Tendency Test

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Measures of Central Tendency Test - 31

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Weekly Quiz Competition

Question 1

1 / -0

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	Number of women
65-68	2
68-71	4
71-74	3
74-77	8
77-80	7
80-83	4
83-86	2

75.92

35.92

60.92

80.92 Solution

No. heart beats per min (Class Interval) ($$f_i$$)	no. of women	$$x_i$$	$$x_if_i$$
65-68	2	66. 5	133
68-71	4	69.5	278
71-74	3	72.5	217.5
74-77	8	75.5	604
77-80	7	78.5	549.5
80-83	4	81.5	326
83-86	2	84.5	169
Total	30		2277

$$Mean\ heart\ beats=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\frac { 2277 }{ 30 } =75.92$$

Question 2
1 / -0

The following table gives the literacy rate (in percentage) of $$35$$ cities. Find the mean literacy rate.
Literacy rate (in $$\%$$)
$$45-55$$
$$55-65$$
$$65-75$$
$$75-85$$
$$85-95$$
No. of cities
$$3$$
$$10$$
$$11$$
$$8$$
$$3$$

A
$$69.43$$

B
$$70.43$$

C
$$68.33$$

D
$$67.23$$

Question 3

1 / -0

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
No. of patients	6	11	21	23	14	5

Find the mean of the data given above.

$$33.42$$ years

$$35.37$$ years

$$37.15$$ years

None of these

Solution

Let us find the mean of the data:

Age (in years)	Number of patients $$f_i$$	Class mark $$x_i$$	$$u_i=\frac {x_i- 30}{10}$$	$$f_i\times u_i$$
5-15	6	10	-2	-12
15-25	11	20	-1	-11
25-35	21	$$30-a$$	0	0
35-45	23	40	1	23
45-55	14	50	2	28
55-65	5	60	3	15
Total	$$n=80$$			43

$$a=30, h=10, n=80$$ and $$\sum f_iu_i=43$$
Mean $$=a+h\times \dfrac {1}{n}\times \sum f_iu_i=30+10\times \dfrac {1}{80}\times 43$$
$$=30+5.37=35.37$$ years
Thus mean $$=35.37$$ years.

Question 4

1 / -0

To find out the concentration of $$SO_2$$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of $$SO_2$$ (in ppm)	Frequency
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

Find the mean concentration of $$SO_2$$ in the air.

0.099

0.090

0.089

0.990 Solution

Concentration of SO2 (in ppm) Class Interval.	No. of localities Frequency(Fi)	Class Mark(Xi)	Fi*Xi
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.1	0.9
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
Total	30		2.96

$$
Mean\quad concentration\quad of\quad { SO }_{ 2 }\quad in\quad air=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\frac { 2.96 }{ 30 } =0.09866\quad =0.099 $$

Question 5

1 / -0

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

No. of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
No. of students	11	10	7	4	4	3	1

12.475

12.405

12.005

12.950 Solution

No. of days (class interval)	No. of students	Classmark($$X_i$$)	$$F_iX_i$$
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39

Total	40		499

We know that Class mark $$ = \dfrac{upper limit +lower limit}{2}$$

Mean number of days $$ = \dfrac{\Sigma F_iX_i}{\Sigma F_i} = \dfrac {499}{40} = 12.475 $$ Days

Question 6

1 / -0

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs.)	No. of households
100-150	4
150-200	5
200-250	12
250-300	2
300-350	2

Find the mean daily expenditure on food by a suitable method.

211

201

215

209 Solution

The mean daily expenditure on food can be calculated using direct method,

Daily [Expenditure]	No. of households [Frequency]	$$x_i$$	$$f_ix_i$$
100-150	4	125	500
150-200	5	175	875
200-250	12	225	2700
250-300	2	275	550
300-350	2	325	650
Total	25		5275

$$\text{Mean daily expenditure on food}=\dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } $$

$$=\dfrac { 5275 }{ 25 } $$

$$=211\\ $$

Question 7

1 / -0

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.)	Number of workers
100-120	12
120-140	14
140-160	8
160-180	6
180-200	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

145.2

155.2

135.2

180.2 Solution

Daily wages (in Rs.) [Class Intervals]	Number of workers [ Fi]	Xi Xi*Fi
100-120	12	110	1320
120-140	14	130	1820
140-160	8	150	1200
160-180	6	170	1020
180-200	10	190	1900
Total	40		5360

$$
Mean\quad daily\quad wage=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\frac { 7260 }{ 50 } =Rs.145.2
$$

Question 8
1 / -0

In a group of students, the mean weight of boys is $$65$$ kg. and mean weight of girls is $$55$$ kg. If the mean weight of all students of group is $$61$$ kg, then the ratio of the number of boys and girls in the group is

A
2 : 3

B
3 : 1

C
3 : 2

D
4 : 3

Solution

Let the no. of boys and girls are $$\displaystyle n_{1}$$ and $$\displaystyle n_{2}$$ respectively.
Mean weight $$=\dfrac{n_1w_1+n_2w_2}{n_1+n_2}$$
$$\Rightarrow \displaystyle \frac{65n_{1}+55n_{2}}{n_{1}+n_{2}}=61$$
$$\Rightarrow 4n_{1}=6n_{2}$$
$$\Rightarrow n_{1}:n_{2}=3:2$$
Question 9
1 / -0

The median of a set of eight numbers is $$4.5$$. Given that seven of the numbers are $$7, 2,3, 4,8,2$$ and $$1$$ find the eighth number and write down the mode of eight numbers.

A
$$4$$

B
$$7$$

C
$$2$$

D
$$1$$

Solution

Let the eighth number be $$x$$.
Then the numbers arranged in ascending order:
$$1,2,2,4,7,8,13,x$$
Since these are $$8$$ in number.
Median value $$\displaystyle =\frac { \left( \frac { 8 }{ 2 } \right) th\quad value+\left( \frac { 8 }{ 2 } +1 \right) th\quad value }{ 2 } $$
$$\displaystyle =\frac { 4th\quad value+5th\quad value }{ 2 } $$
$$\displaystyle =\frac { 4+7 }{ 2 } =\frac { 11 }{ 2 } =5.5$$
$$\displaystyle \therefore $$ Mode $$= 2$$
Question 10
1 / -0

The mean of $$1, 7, 5, 3, 4$$ and $$4$$ is $$m$$. The observations $$3, 2, 4, 2, 3, 3$$ and $$p$$ have the mean $$(m - 1)$$. Find the median of the second set of data.

A
$$4$$

B
$$2.5$$

C
$$3$$

D
$$5$$

Solution

Mean $$=\displaystyle m=\frac { 1+7+5+3+4+4 }{ 6 } =\frac { 24 }{ 6 } =4$$
Now mean of $$3,2, 4, 2, 3, 3$$ and $$p$$ is $$(m-1)=(4 -1) = 3$$
$$\displaystyle \therefore \frac { 3+2+4+2+3+3+p }{ 7 } =3$$
$$\displaystyle \Rightarrow \frac { 17+p }{ 7 } =3\Rightarrow 17+p=21\Rightarrow p=4$$
$$\displaystyle \therefore $$ The observations are $$3, 2, 4, 2, 3, 3$$ and $$4$$.
Arranged in ascending order, the numbers are $$2,2,3,3,3,4,4$$.
Median $$\displaystyle =\left( \frac { 7+1 }{ 2 } \right)^{th} \:\text{value}= 4^{th} \:\text{value}= 3$$

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Re-Attempt Weekly Quiz Competition

Literacy rate (in $$\%$$)	$$45-55$$	$$55-65$$	$$65-75$$	$$75-85$$	$$85-95$$
No. of cities	$$3$$	$$10$$	$$11$$	$$8$$	$$3$$

Measures of Central Tendency Test - 31

Result

Measures of Central Tendency Test - 31

Score

-

Rank

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SHARING IS CARING

Question 1

75.92

35.92

60.92

80.92

Solution

Question 2

$$69.43$$

$$70.43$$

$$68.33$$

$$67.23$$

Question 3

$$33.42$$ years

$$35.37$$ years

$$37.15$$ years

None of these

Solution

Question 4

0.099

0.090

0.089

0.990

Solution

Question 5

12.475

12.405

12.005

12.950

Solution

Question 6

211

201

215

209

Solution

Question 7

145.2

155.2

135.2

180.2

Solution

Question 8

2 : 3

3 : 1

3 : 2

4 : 3

Solution

Question 9

$$4$$

$$7$$

$$2$$

$$1$$

Solution

Question 10

$$4$$

$$2.5$$

$$3$$

$$5$$

Solution

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