Statistics Test

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Statistics Test - 43

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Weekly Quiz Competition

Question 1
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 2

1 / -0

Find the median for the following data given below$$:-$$

Class interval	$$11-21$$	$$21-31$$	$$31-41$$	$$41-51$$
Frequencies	$$1$$	$$5$$	$$7$$	$$3$$

$$33.85$$

$$45.28$$

$$55.28$$

$$65.28$$

Solution

Class interval	Frequencies(f)	Cumulative frequencies(cf)
$$11-21$$	$$1$$	$$1$$
$$21-31$$	$$5$$	$$1+5=6$$
$$31-41$$	$$7$$	$$6+7=13$$
$$41-51$$	$$3$$	$$13+3=16$$
	$$n=\Sigma f=16$$ $$\displaystyle\frac{n}{2}=\frac{16}{2}=8$$

Median $$=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$$
L$$=$$ Lower limit of the median class $$=31$$
$$n=\Sigma f=$$ Total number of frequency $$=16$$
$$\displaystyle\frac{n}{2}=\frac{16}{2}=8$$
So, the median class is $$31-41$$
cf$$=$$ Cumulative frequency of the class preceding the median class $$=6$$
f$$=$$ Frequency of the median class $$=7$$
h$$=$$ class width$$=10$$
Therefore, Median $$=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$$
$$=31+\left(\displaystyle\frac{\frac{16}{2}-6}{7}\right)\times 10$$
$$=31+\left(\displaystyle\frac{8-6}{7}\right)\times 10$$
$$=31+4.28$$
$$=35.28$$
Hence, the median is $$35.28$$

Question 3

1 / -0

Calculate the median for the following data given below$$:$$

X	$$8-10$$	$$10-12$$	$$12-14$$	$$14-16$$
F	$$4$$	$$8$$	$$12$$	$$16$$

$$13.33$$

$$11.33$$

$$12.33$$

$$10.33$$

Solution

X	Frequencies (f)	Cumulative frequency (cf)
$$8-10$$	$$4$$	$$4$$
$$10-12$$	$$8$$	$$4+8=12$$
$$12-14$$	$$12$$	$$12+12=24$$
$$14-16$$	$$16$$	$$24+16=40$$

Here,
$$n=\Sigma f=$$ Total frequency $$=4+8+12+16=40$$
$$\displaystyle\frac{n}{2}=\frac{40}{2}=20$$
$$20^{th}$$ observation lies between cumulative frequencies $$12$$ and $$24$$

So, the median class is $$12-14$$

We know, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$l=$$ Lower limit of the median class $$=12$$
$$cf=$$ Cumulative frequency of the class preceding the median class $$=12$$
$$f=$$ Frequency of the median class $$=12$$
$$h=$$ class width$$=14-12=2$$

Therefore,

Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=12+\left(\displaystyle\frac{\displaystyle\frac{40}{2}-12}{12}\right)\times 2$$
$$=12+\left(\displaystyle\frac{20-12}{12}\right)\times 2$$
$$=12+1.333$$
$$=13.33$$

Hence, the median is $$13.33$$.

Question 4
1 / -0

Calculate the median for the following data given below$$:$$
X
$$5-9$$
$$9-13$$
$$13-17$$
$$17-21$$
F
$$3$$
$$8$$
$$9$$
$$4$$

A
$$11.44$$

B
$$12.44$$

C
$$13.44$$

D
$$15.44$$

Solution

X
Frequencies (f)
Cumulative frequency (cf)
$$5-9$$
$$3$$
$$3$$
$$9-13$$
$$8$$
$$3+8=11$$
$$13-17$$
$$9$$
$$11+9=20$$
$$17-21$$
$$4$$
$$20+4=24$$

$$n=\Sigma f=24$$
$$\displaystyle\frac{n}{2}=\frac{24}{2}=12$$

Median $$=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ Total number of frequency $$=12$$
$$\displaystyle\frac{n}{2}=\frac{24}{2}=12$$
So, the median class is $$13-17$$
L$$=$$ Lower limit of the median class $$=13$$
cf$$=$$ Cumulative frequency of the class preceding the median class $$=11$$
f$$=$$ Frequency of the median class $$=9$$
h$$=$$ class width$$=4$$
Therefore, Median $$=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$$
$$=13+\left(\displaystyle\frac{\frac{24}{2}-11}{9}\right)\times 4$$
$$=13+\left(\displaystyle\frac{12-11}{9}\right)\times 4$$
$$=13+0.44$$
$$=13.44$$
Hence, the median is $$13.44$$.

Question 5

1 / -0

Estimate the median for the following data given below$$:$$

Marks	$$50-55$$	$$55-60$$	$$60-65$$	$$65-70$$	$$70-75$$
Students	$$50$$	$$40$$	$$30$$	$$20$$	$$10$$

$$53.25$$

$$55.25$$

$$58.125$$

$$63.125$$

Solution

Marks	Students (f)	Cumulative frequency (cf)
$$50-55$$	$$50$$	$$50$$
$$55-60$$	$$40$$	$$50+40=90$$
$$60-65$$	$$30$$	$$90+30=120$$
$$65-70$$	$$20$$	$$120+20=140$$
$$70-75$$	$$10$$	$$140+10=150$$

Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ Total number of frequency $$=150$$
$$\displaystyle\frac{n}{2}=\frac{150}{2}=75$$
So, the median class is $$55-60$$
L$$=$$ Lower limit of the median class $$=55$$
cf$$=$$ Cumulative frequency of the class preceding the median class $$=50$$
f$$=$$ Frequency of the median class $$=40$$
h$$=$$ class width$$=5$$
Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=55+\left(\displaystyle\frac{\displaystyle\frac{150}{2}-50}{40}\right)\times 5$$
$$=55+\left(\displaystyle\frac{75-50}{40}\right)\times 5$$
$$=55+\dfrac{25}{8}$$
$$=58.125$$
Hence, the median is $$58.125$$.

Question 6
1 / -0

Find the median for the following data given below$$:$$
X
$$0-5$$
$$5-10$$
$$10-15$$
$$15-20$$
$$20-25$$
F
$$4$$.
$$8$$
$$12$$
$$16$$
$$20$$

A
$$16.875$$

B
$$15.875$$

C
$$14.875$$

D
$$13.875$$
Question 7
1 / -0

Calculate the median for the following data given below$$:$$
Scores
$$12-15$$
$$15-18$$
$$18-21$$
$$21-24$$
Frequency
$$10$$
$$20$$
$$30$$
$$40$$

A
$$14$$

B
$$15$$

C
$$16$$

D
$$17$$
Question 8
1 / -0

Find the median for the following data given below:
Class interval
$$0-2$$
$$2-4$$
$$4-6$$
$$6-8$$
Frequencies
$$2$$
$$3$$
$$5$$
$$7$$

A
$$5.3$$

B
$$5.4$$

C
$$5.1$$

D
$$5.6$$

Solution

Class Interval Frequency(f) Cumulative Frequency (cf)
0-2 2 2
2-4 3 5
4-6 5 10
6-8 7 17
$$n=\Sigma f=17$$

Here, $$\displaystyle\frac{n}{2}=\frac{17}{2}=8.5$$

Hence, the median class is $$4-6$$
We know that,
Median $$=l+\left(\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
Where, $$l\rightarrow$$ lower limit of the median class $$=4$$
$$n=\Sigma f\rightarrow$$ total number of frequency $$=17$$
$$cf\rightarrow$$ cumulative frequency of the class preceding the median class $$=5$$
$$f\rightarrow$$ frequency of the median class$$=5$$
$$h\rightarrow$$ class width $$=2$$
Therefore, Median $$=4+\left(\displaystyle\frac{\frac{17}{2}-5}{5}\right)\times 2$$
$$\Rightarrow 4+\left(\displaystyle\frac{17-10}{2\times 5}\right)\times 2$$
$$\Rightarrow 4+1.4$$
$$\Rightarrow 5.4$$
Hence, the median for the given data is $$5.4$$
Question 9
1 / -0

Find the median for the following data given below$$:-$$
X
$$4-7$$
$$7-10$$
$$10-13$$
$$13-14$$
F
$$2$$
$$5$$
$$3$$
$$2$$

A
$$11$$

B
$$10$$

C
$$9$$

D
$$8$$

Solution

X
F(f)
Cumulative frequency (cf)
$$4-7$$
$$2$$
$$2$$
$$7-10$$
$$5$$
$$2+5=7$$
$$10-13$$
$$3$$
$$7+3=10$$
$$13-14$$
$$2$$
$$10+2=12$$

$$n=\Sigma f=12$$
$$\dfrac{n}{2}$$$$=\dfrac{12}{2}=6$$

Median $$=l+\left(\displaystyle\frac{\frac{n}{2}-cf}{f}\right)\times h$$
$$n=\Sigma f=$$ total number of frequency $$=12$$
$$\displaystyle\frac{n}{2}=\frac{12}{2}=6$$

So, the median class is $$10-13$$
$$L=$$ Lower limit of median class $$=10$$
$$cf=$$ Cumulative frequency of the class preceding the median class $$=7$$
$$f=$$ frequency of the median class $$=3$$
$$h=$$ class width $$=3$$
Therefore, Median $$=l+\left(\displaystyle\frac{\displaystyle\frac{n}{2}-cf}{f}\right)\times h$$
$$=10+\left(\displaystyle\frac{\displaystyle\frac{12}{2}-7}{3}\right)\times 3$$
$$=10+\left(\displaystyle\frac{6-7}{3}\right)\times 3$$
$$=10-1$$
$$=9$$
Hence, the median is $$9$$
Question 10
1 / -0

The marks obtained by $$20$$ students in maths are $$35, 40, 78, 30, 50, 78, 36, 65, 78, 40, 30, 65, 50, 35, 80, 95, 80, 36, 95, 35$$. Find the Arithmetic mean of the students by direct method.

A
$$85.56$$

B
$$65.86$$

C
$$56.85$$

D
$$86.56$$

Solution

Arranging in ascending order
$$30, 30, 35, 35, 35, 36, 36, 40, 40, 50, 50, 65, 65, 78, 78, 78, 80, 80, 95, 95$$.

$$X_i$$ $$F_i$$ $$X_i F_i$$
30 2 60
35 3 105
36 2 72
40 2 80
50 2 100
65 2 130
78 3 156
80 2 160
95 2 190
$$\Sigma F_i = 20$$ $$\Sigma F_i X_i=1131$$
Mean = $$\cfrac{\Sigma F_i X_i}{\Sigma F_i} = \cfrac{1131}{20} = 56.55 \simeq 56.85$$
C) 56.85

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X	Frequencies (f)	Cumulative frequency (cf)
$$5-9$$	$$3$$	$$3$$
$$9-13$$	$$8$$	$$3+8=11$$
$$13-17$$	$$9$$	$$11+9=20$$
$$17-21$$	$$4$$	$$20+4=24$$
	$$n=\Sigma f=24$$ $$\displaystyle\frac{n}{2}=\frac{24}{2}=12$$

X	$$0-5$$	$$5-10$$	$$10-15$$	$$15-20$$	$$20-25$$
F	$$4$$.	$$8$$	$$12$$	$$16$$	$$20$$

Scores	$$12-15$$	$$15-18$$	$$18-21$$	$$21-24$$
Frequency	$$10$$	$$20$$	$$30$$	$$40$$

Class interval	$$0-2$$	$$2-4$$	$$4-6$$	$$6-8$$
Frequencies	$$2$$	$$3$$	$$5$$	$$7$$

Class Interval	Frequency(f)	Cumulative Frequency (cf)
0-2	2	2
2-4	3	5
4-6	5	10
6-8	7	17
	$$n=\Sigma f=17$$

X	F(f)	Cumulative frequency (cf)
$$4-7$$	$$2$$	$$2$$
$$7-10$$	$$5$$	$$2+5=7$$
$$10-13$$	$$3$$	$$7+3=10$$
$$13-14$$	$$2$$	$$10+2=12$$
	$$n=\Sigma f=12$$ $$\dfrac{n}{2}$$$$=\dfrac{12}{2}=6$$

$$X_i$$	$$F_i$$	$$X_i F_i$$
30	2	60
35	3	105
36	2	72
40	2	80
50	2	100
65	2	130
78	3	156
80	2	160
95	2	190
	$$\Sigma F_i = 20$$	$$\Sigma F_i X_i=1131$$

Statistics Test - 43

Result

Statistics Test - 43

Score

-

Rank

-

SHARING IS CARING

Question 1

$$9.394 $$ cm

$$9.925 $$ cm

$$7.925 $$ cm

$$6.925 $$ cm

Solution

Question 2

$$33.85$$

$$45.28$$

$$55.28$$

$$65.28$$

Solution

Question 3

$$13.33$$

$$11.33$$

$$12.33$$

$$10.33$$

Solution

Question 4

$$11.44$$

$$12.44$$

$$13.44$$

$$15.44$$

Solution

Question 5

$$53.25$$

$$55.25$$

$$58.125$$

$$63.125$$

Solution

Question 6

$$16.875$$

$$15.875$$

$$14.875$$

$$13.875$$

Question 7

$$14$$

$$15$$

$$16$$

$$17$$

Question 8

$$5.3$$

$$5.4$$

$$5.1$$

$$5.6$$

Solution

Question 9

$$11$$

$$10$$

$$9$$

$$8$$

Solution

Question 10

$$85.56$$

$$65.86$$

$$56.85$$

$$86.56$$

Solution

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