Measures of Central Tendency Test

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Measures of Central Tendency Test - 36

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Weekly Quiz Competition

Question 1

1 / -0

Number of dog's	2-4	4-6	6-8	8-10
weight (kg)	20	10	12	15

Find the arithmetic mean of dog's weight using direct method.

$$5.17$$ kg

$$4.56$$ kg

$$5.77$$ kg

$$4.58$$ kg

Solution

Number of dog's	Weight (kg) $$(f_i)$$	Midpoint $$(X_i)$$	$$f_iX_i$$
2-4	20	3	$$20\times 3=60$$
4-6	10	5	$$10\times 5=50$$
6-8	12	7	$$12\times 7=84$$
8-10	15	9	$$15\times 9=135$$
	$$\sum f_i = 57$$		$$\sum f_ix_i = 329$$

Arithmetic mean using direct method formula is $$\bar { x } = \dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } $$
$$\bar { x } = \dfrac { 329 }{ 57 } $$
$$\bar { x } = 5.77$$
The average mean weight of the dog is $$5.77 kg$$.

Question 2
1 / -0

Find the mean of the above data.

A
$$90$$

B
$$91$$

C
$$94$$

D
$$92$$

Solution

Answer:- Using the direct method

Marks obtained
$$x_i$$ No. of students
$$f_i$$ $$f_i x_i$$
100 4 400
89 5 445
98 8 784
87 2 174
75 1 75
95 3 285
$$\sum f_i = 23$$ $$\sum f_i x_i = 2167$$
$$Mean=\cfrac{\sum f_i x_i}{\sum f_i} $$
$$= \cfrac{2167}{23}$$
$$ = 94.21$$
$$ \approx 94$$

Question 3

1 / -0

Walmart supermarket recorded the length of time, to the nearest hour, that a sample of $$100$$ cars was parked in their lot. Estimate the average hour using direct method.

$$20$$ hours

$$31$$ hours

$$25$$ hours

$$27$$ hours

Solution

Answer:-

Representing the data into frequency distribution table.

Daily Pocket allowance	No. of workers	$$x_i$$	$$f_i x_i$$
5-15	20	10	200
15-25	30	20	600
25-35	15	30	450
35-45	12	40	480
45-55	6	50	300
55-65	17	60	1020

	$$\Sigma f_i=100$$		$$\Sigma f_i x_i = 3050$$

$$\therefore \; Mean=\cfrac{\Sigma f_i x_i}{\Sigma f_i}=\cfrac{3050}{100}=30.5 \simeq 31$$

The average hours are 31.

Question 4
1 / -0

Find the arithmetic mean of the children's height using direct method.

A
$$9.394 $$ cm

B
$$9.925 $$ cm

C
$$7.925 $$ cm

D
$$6.925 $$ cm

Solution

Arithmetic mean using direct method formula is $$\bar { x } = \dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } $$
$$\bar { x } = \dfrac { 1785 }{ 190 } $$
$$\bar { x } = 9.3947$$
The average mean height of the children is $$9.39$$ cm.
Question 5
1 / -0

A group of $$50$$ house owners contributes money towards children's education of their street. The amount of money collected is shown in the table below: (use direct method).

A
Rs. $$27$$

B
Rs. $$17$$

C
Rs. $$10$$

D
Rs. $$23$$

Solution

Answer:- Given $$\Sigma f=50$$
Using frequency distribution table

Amount No. of house owners $$x_i$$ $$x_i f_i$$
0-10 5 5 25
10-20 10 15 150
20-30 15 25 375
30-40 10 35 350
40-50 10 45 450
$$\Sigma f_i=50$$ $$\Sigma f_i x_i = 1350$$
$$\therefore$$ Amount of money collected = Arithmetic mean
$$\Rightarrow \; Mean = \cfrac{\Sigma f_i x_i}{\Sigma f_i} = \cfrac{1350}{50}=27$$
$$\therefore \; $$ the amount of money collected = 27Rs.
A) Rs. 27

Question 6

1 / -0

The fruit vendor has apples packed in boxes. Each box contains different numbers of apples. The following was the distribution of apples according to the number of boxes. Find the mean number of Apples kept in a packing box using the direct method.

$$39.5$$

$$41.5$$

$$38.5$$

$$35.5$$

Solution

No. of apples	No. of boxes $$f_i$$	$$x_i = \cfrac{\text{Upper limit+Lower limit}}{2}$$	$$f_i x_i $$
30-33	60	31.5	1890
33-36	120	34.5	4140
36-39	150	37.5	5625
39-42	80	40.5	3160
42-45	50	43.5	2175
45-48	90	46.5	4185
	$$\Sigma f_i = 550$$		$$\Sigma f_i x_i = 21175$$

By direct method

$$\text{Mean} = \cfrac{\Sigma f_i x_i}{\Sigma f_i} = \cfrac{21175}{550} = 38.5$$

$$\therefore$$ Mean $$=$$ $$38.5$$

Question 7
1 / -0

$$30$$ people gave money for donation. Find the average amount of money collected. (use Assumed mean).

A
$$150$$

B
$$180$$

C
$$170$$

D
$$175$$

Solution

Money X People
(F) d=X-A Fd
75-125 100 7 -50 -350
125-175 150=A 8 0 0
175-225 200 8 100 800
225-275 250 7 150 1050
$$\Sigma F = 30$$ $$\Sigma Fd = 1500$$
Mean (By shortcut method) = $$A+\cfrac{\Sigma Fd}{\Sigma F} = 100+\cfrac{1500}{30} = 100+50 = 150$$
D) Mean = 150

Question 8

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Class interval	$$10-15$$	$$15-20$$	$$20-25$$	$$25-30$$
Frequency	$$7$$	$$12$$	$$8$$	$$10$$

Find the mean of the following data using Assumed mean method.

$$19.781$$

$$20.836$$

$$20.337$$

$$21.005$$

Solution

Answer:- Using Shortcut method

Class Interval	$$x $$	Frequency $$f $$	$$d=x-A $$	$$f_d$$
$$10-15 $$	$$13 $$	$$7$$	$$-4$$	$$-28$$
$$15-20$$	$$17=A$$	$$12$$	$$0$$	$$0$$
$$20-25$$	$$23$$	$$8$$	$$6$$	$$48$$
$$25-30$$	$$27$$	$$10$$	$$10 $$	$$100$$
		$$\Sigma f=37$$		$$\Sigma f_d = 120$$

Mean = $$A + \cfrac{\Sigma f_d}{\Sigma f} = 17+\cfrac{120}{37} = 17 + 3.337 = 20.337$$

Hence option $$(C)$$ is correct

Question 9
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The marks obtained by $$20$$ students of Class $$X$$ of a certain school in a English paper consisting of $$100$$ marks are presented in table below. Find the mean of the marks obtained by the students using step deviation method.

A
$$61$$

B
$$62$$

C
$$63$$

D
$$64$$

Solution

Answer:- Using Shortcut Method
Class interval width (i) = $$50-40 = 10$$

Marks X F $$d=\cfrac{X-A}{i}$$ Fd
40-50 45 5 -2 -10
50-60 55 4 -1 -4
60-70 65 = A 3 0 0
70-80 75 6 1 6
80-90 85 2 2 4
$$\Sigma F = 20$$ $$\Sigma Fd = -4$$
Mean ( By step deviation formula) = $$A+\cfrac{\Sigma Fd}{\Sigma F} \times i = 65+\left(\cfrac{-4}{20} \times 10 \right) = 65=2 = 63$$
C) Mean = $$63$$

Question 10

1 / -0

What is the average weight of $$40$$ people whom are diagnosed for thyroid? (use Assumed mean method).

$$90.75$$

$$101.45$$

$$87.25$$

$$95$$

Solution

Weight	X = $$\cfrac{U.L.+L.L.}{2}$$	Patients (F)	d=X-A	Fd
65-75	70	3	-30	-30
75-85	80	13	-20	-260
85-95	90	12	-10	-120
95-105	100 = 100	8	0	0
105-115	110	4	10	40
		$$\Sigma F = 40$$		$$\Sigma Fd = -430$$

Mean by Shortcut Method = $$A+\cfrac{\Sigma Fd}{\Sigma F} = 100+\cfrac{-370}{40} = 100-9.25 = 90.75$$

A) Mean = 90.75

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Marks obtained $$x_i$$	No. of students $$f_i$$	$$f_i x_i$$
100	4	400
89	5	445
98	8	784
87	2	174
75	1	75
95	3	285
	$$\sum f_i = 23$$	$$\sum f_i x_i = 2167$$

Amount	No. of house owners	$$x_i$$	$$x_i f_i$$
0-10	5	5	25
10-20	10	15	150
20-30	15	25	375
30-40	10	35	350
40-50	10	45	450
	$$\Sigma f_i=50$$		$$\Sigma f_i x_i = 1350$$

Money	X	People (F)	d=X-A	Fd
75-125	100	7	-50	-350
125-175	150=A	8	0	0
175-225	200	8	100	800
225-275	250	7	150	1050
		$$\Sigma F = 30$$		$$\Sigma Fd = 1500$$

Marks	X	F	$$d=\cfrac{X-A}{i}$$	Fd
40-50	45	5	-2	-10
50-60	55	4	-1	-4
60-70	65 = A	3	0	0
70-80	75	6	1	6
80-90	85	2	2	4
		$$\Sigma F = 20$$		$$\Sigma Fd = -4$$

Measures of Central Tendency Test - 36

Result

Measures of Central Tendency Test - 36

Score

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Rank

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SHARING IS CARING

Question 1

$$5.17$$ kg

$$4.56$$ kg

$$5.77$$ kg

$$4.58$$ kg

Solution

Question 2

$$90$$

$$91$$

$$94$$

$$92$$

Solution

Question 3

$$20$$ hours

$$31$$ hours

$$25$$ hours

$$27$$ hours

Solution

Question 4

$$9.394 $$ cm

$$9.925 $$ cm

$$7.925 $$ cm

$$6.925 $$ cm

Solution

Question 5

Rs. $$27$$

Rs. $$17$$

Rs. $$10$$

Rs. $$23$$

Solution

Question 6

$$39.5$$

$$41.5$$

$$38.5$$

$$35.5$$

Solution

Question 7

$$150$$

$$180$$

$$170$$

$$175$$

Solution

Question 8

$$19.781$$

$$20.836$$

$$20.337$$

$$21.005$$

Solution

Question 9

$$61$$

$$62$$

$$63$$

$$64$$

Solution

Question 10

$$90.75$$

$$101.45$$

$$87.25$$

$$95$$

Solution

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