Descriptive Statistics Test 34

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Descriptive Statistics Test 34

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Weekly Quiz Competition

Question 1

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For the information given below,Find quartile deviation and coefficient of quartile deviation

Maximum Load	Number of Cables
9.3-9.7	22
9.8-10.2	55
10.3-10.7	12
10.8-11.2	17
11.3-11.7	14
11.8-12.2	66
12.3-12.7	33
12.8-13.2	11

$$0.275,0.065$$

$$0.495,0.045$$

$$0.415,0.055$$

$$0.225,0.012$$

Solution

$$Maximum\,load$$	$$No.\,of\,cables$$	$$Class\,boundaries$$	$$Cumulative\,frequency$$
$$9.3-9.7$$	$$2$$	$$9.25-9.75$$	$$2$$
$$9.8-10.2$$	$$5$$	$$9.75-10.25$$	$$7$$
$$10.3-10.7$$	$$12$$	$$10.25-10.75$$	$$19$$
$$10.8-11.2$$	$$17$$	$$10.75-11.25$$	$$36$$
$$11.3-11.7$$	$$14$$	$$11.25-11.75$$	$$50$$
$$11.8-12.2$$	$$6$$	$$11.75-12.25$$	$$56$$
$$12.3-12.7$$	$$3$$	$$12.25-12.75$$	$$59$$
$$12.8-13.2$$	$$1$$	$$13.25$$	$$60$$

$$Q_1:$$

Value of $$\dfrac{n}{4}^{th}=\dfrac{60}{4}=15^{th}$$

$$\therefore$$ $$Q_1$$ lies in class $$10.25-10.75$$

Where, $$l=10.25,\,h=0.5,\,f=12,\,cf=7$$

$$Q_1=l+\dfrac{h}{f}\left(\dfrac{n}{4}-c.f.\right)$$

$$=10.25+\dfrac{0.5}{12}(15-7)$$

$$=10.25+0.33$$

$$=10.58$$

$$Q_3:$$

Value of $$\dfrac{3n}{4}^{th}item=\dfrac{3\times 60}{4}=45^{th}item$$

$$\therefore$$ $$Q_3$$ lies in class $$11.25-11.75$$

Here, $$l=11.25,\,h=0.5,\,f=14,\,\dfrac{3n}{4}=45$$ and $$cf=36$$

$$Q_3=l+\dfrac{h}{f}\left(\dfrac{3n}{4}-c.f\right)$$

$$=11.25+\dfrac{0.5}{14}(45-36)$$

$$=11.25+0.32$$

$$=11.57$$

$$\Rightarrow$$ Now, Quartile deviation $$=\dfrac{Q_3-Q_1}{2}$$

$$=\dfrac{11.57-10.58}{2}$$

$$=\dfrac{0.99}{2}$$

$$=0.495$$

$$\Rightarrow$$ Coefficient of Quartile Deviation $$=\dfrac{Q_3-Q_1}{Q_3+Q_1}$$

$$=\dfrac{11.57-10.58}{11.57+10.58}$$

$$=\dfrac{0.99}{22.15}$$

$$=0.045$$

Question 2
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Mean of $$10$$ observations is $$50$$ and their standard deviation is $$10$$. If each observation is subtracted by $$5$$ and then divided by $$4$$, then the new mean and standard deviation are

A
$$22.45, 2.5$$

B
$$11.25, 2.5$$

C
$$11.5, 2.5$$

D
$$11, 2.5$$

E
$$11.75, 2.5$$

Solution

Given,mean of $$10$$ observations $$= \dfrac {\displaystyle \sum_{i = 1}^{10} x_{i}}{10} = 50$$
$$\Rightarrow \displaystyle \sum_{i = 1}^{10} x_{i} = 500$$ .... (i)
Therefore, new mean $$= \dfrac {\displaystyle \sum_{i = 1}^{10} x_{i} - 5\times 10}{4\times 10} = \dfrac {500 - 50}{4\times 10}$$ [using Eq. (i)]
$$= \dfrac {450}{4\times 10} = 11.25$$
and standard deviation $$= \dfrac {\text{old}(SD)}{4} = \dfrac {10}{4} = 2.5$$.
Question 3
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The arithmetic mean of the observations 10,8,5,a,b is 6 and their variance 6.8. Then ab=

A
6

B
4

C
3

D
12

Solution

$$\dfrac{10+8+5+a+b}{5}=6,$$

$$\dfrac{1}{n}\sum \left ( x_i-\overline{x} \right )^{2}=6.8$$
solving above two equations we get values of a and b.
Question 4
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Directions For Questions

The graph gives expenditure of a company in the years $$2003,2004$$ and $$2005$$ for the month January to July. Read the graph and answer the questions.

...view full instructions

What is the average monthly expenditure (Rs. in lakhs) for January to July during the year 2005?

A
$$658.3$$

B
$$766.7$$

C
$$764.2$$

D
$$657.1$$

Solution

Required average
$$=\cfrac { 700+750+850+850+600+750+850 }{ 7 } $$
$$=\cfrac { 5350 }{ 7 } =764.2$$ lakh
Question 5
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The line graph show the number of cars Jatin sold over the past $$5$$ weeks.
If he was paid $$Rs \ 20000$$ for every car sold, how much was he paid over the past $$5$$ weeks?

A
$$Rs \ 280000$$

B
$$Rs \ 240000$$

C
$$Rs \ 480000$$

D
$$Rs \ 500000$$

Solution

Total number of cars sold by jatin in week $$1=2$$
Total number of cars sold by jatin in week $$2=4$$
Total number of cars sold by jatin in week $$3=3$$
Total number of cars sold by jatin in week $$4=6$$
Total number of cars sold by jatin in week $$5=9$$
Therefore Total numbers of cars sold upto $$5$$ weeks =$$2+4+3+6+9=24$$
For every caar sold , he paid $$20000$$ Rs
So for 24 cars he would have paid=$$20000 \times 24= 480000$$.
Question 6
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An Um contains $$4$$ white and $$3$$ red balls. $$X$$ is the no. of red balls. Find the Mean and Variance of $$X$$.

A
$$\dfrac {9}{7}, \dfrac {24}{49}$$

B
$$\dfrac {9}{7}, \dfrac {25}{48}$$

C
$$\dfrac {9}{8}, \dfrac {24}{49}$$

D
$$\dfrac {9}{8}, \dfrac {24}{48}$$

Solution

Let x be the no. of red balls in a random draw of three balls. As these are $$8$$ red balls, double value of x are $$0, 1, 2, 3$$
$$p(0)=\dfrac{^{3}C_o\times {^{4}C_3}}{^{7}C_3}=\dfrac{4\times 3\times 2}{7\times 6\times 5}=\dfrac{4}{35}$$
$$p(1)=\dfrac{^{3}C_1\times {^{4}C_2}}{^{7}C_3}=\dfrac{3\times 6\times 6}{7\times 6\times 5}=\dfrac{18}{35}$$
$$p(2)=\dfrac{^{3}C_2\times {^{4}C_1}}{^{7}C_3}=\dfrac{3\times 4\times 6}{7\times 6\times 5}=\dfrac{12}{35}$$
$$p(3)=\dfrac{^{3}C_3\times {^{4}C_3}}{^{7}C_3}=\dfrac{1\times 1\times 6}{7\times 6\times 5}=\dfrac{1}{35}$$
for calculation of mean of variance
x    p(x) xp(x)    $$x^2p(x)$$
$$0$$ $$4/35$$    $$0$$    $$0$$
$$1$$    $$18/35$$ $$18/35$$    $$18/35$$
$$2$$ $$12/35$$ $$24/35$$ $$40/35$$
$$3$$ $$1/35$$ $$3/35$$    $$9/35$$
Total    $$1$$ $$9/7$$ $$15/7$$
Mean $$=\sum x p(x)=\dfrac{9}{7}$$
variance$$=\sum x^2\cdot p(x)-(\sum x.p(x))^2=\dfrac{15}{7}-\dfrac{81}{49}=\dfrac{24}{49}$$.
Question 7
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The given pictograph shows the number of wall clocks sold by a company during a week..Each clock in the pictograph represents $$5$$ wall clocks. How many more wall clocks were sold on Thursday than on Tuesday?

A
$$9$$

B
$$8$$

C
$$5$$

D
$$10$$

Solution

Number of wall clocks sold on Tuesday $$5$$ $$\times$$ $$5 = 25$$
Number of wall clocks sold on Thursday = $$5$$ $$\times$$ $$7 = 35$$
$$\therefore$$ Required difference = $$35 - 25 = 10$$
Question 8
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The given bar graph shows the number of residents at 6 places. Study the graph and answer the questions. How many more residents were residing at China town than at Lajpat nagar?

A
$$1600$$

B
$$1800$$

C
$$1400$$

D
$$1200$$

Solution

Number of residents at China town = $$6000$$
Number of residents at Lajpat nagar = $$4400$$
Required difference = $$6000-4400=1600$$
Question 9
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The bar graph shows the number of traffic accidents in a country from year $$2000$$ to $$2014$$ . What is the difference between accidents in the year $$2001$$ and $$2004$$.

A
$$60,000$$

B
$$40,000$$

C
$$20,000$$

D
$$45,000$$

Solution

The letter $$k$$ is used to denote $$1000$$

Therefore, Number of traffic accidents in the year $$2001$$ $$=260k$$

$$=260\times1000$$

$$=260,000$$

Number of traffic accidents in year $$2004$$ $$=220k$$

$$=220\times1000$$

$$=220,000$$.

Then, the difference $$=260,000-220,000$$

$$=40,000$$

Therefore, option $$B$$ is correct.
Question 10
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The given pictograph shows the number of wall clocks sold by a company during a week.Each clock in the pictograph represents $$5$$ wall clocks. On what day was the sale maximum?

A
Monday

B
Tuesday

C
Saturday

D
Thursday

Solution

As shown in the figure, the sale was maximum on Thursday.
Hence the correct answer is option D