Measures of Central Tendency Test

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Measures of Central Tendency Test - 19

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Question 1
1 / -0

Sam's test scores are History $$76$$, Geography $$74$$, Math $$92$$, English $$81$$ and Chemistry $$80$$. If the average (arithmetic mean) score is $$M$$ and the median score is $$m$$, what is the value of $$M-m$$?

A
0.4

B
0.5

C
0.6

D
0.8

Solution

Score in History $$76$$, Geography $$74$$, Math $$92$$, English $$81$$, Chemistry $$80$$
$$\therefore$$ average Sore $$M=$$ $$\dfrac{76+74+92+81+81}{5}=\dfrac{403}{5}=80.6$$
Score $$=74,76,80,81,92$$
$$\therefore$$ Median $$(m)=80$$
$$\Rightarrow M-m=80-80.6=0.6$$
Question 2
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The mean of $$a, b, c, d$$ and $$e$$ is $$28$$. If the mean of $$a, c$$ and $$e$$ is $$24$$, what is the mean of $$b$$ and $$d$$?

A
$$31$$

B
$$32$$

C
$$33$$

D
$$34$$

Solution

Formula used:
$$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$$

Given $$\Rightarrow$$ $$\dfrac{a+b+c+d+e}{5}=28$$ and $$\dfrac{a+c+e}{3} =24 $$
To find $$\Rightarrow$$ $$\dfrac{b+d}{2}$$ which is nothing but the mean of $$b$$ and $$d$$.

$$\begin{aligned}{}\dfrac{{a + b + c + d + e}}{5} &= 28\\a + b + c + d + e& = 140\quad \dots (i)\end{aligned}$$

Also,
$$\begin{aligned}{}\dfrac{{a + c + e}}{3} &= 24\\a + c + e& = 72\quad \dots (ii)\end{aligned}$$

By using $$(i)$$ and $$(ii)$$,
$$\begin{aligned}{}a + b + c + d + e &= 140\\b + d + 72 &= 140\\b + d &= 68\\\dfrac{{b + d}}{2} &= 34\end{aligned}$$

Hence, option $$D$$ is the correct answer.
Question 3
1 / -0

In which two measure of central tendency do not affect the outliers?

A
median and mean

B
mean and mode

C
median and mode

D
mode and range

Solution

Median and mode are the two measure of central tendency do not affect the outliers.
Question 4
1 / -0

What is the mode of $$2, 7, 8, 11, 7, 1, 9, 8, 7$$?

A
$$9$$

B
$$8$$

C
$$7.5$$

D
$$7$$

Solution

$$2,7,8,11,7,1,9,8,7$$
As frequency of $$7$$ is maximum $$=3$$
$$\therefore $$ $$\boxed{Mode\, is\, 7}$$
Question 5
1 / -0

If the median of $$\dfrac {x}{3}, \dfrac {x}{2}, \dfrac {x}{4}, \dfrac {2x}{9}$$ and $$x$$ is $$5$$, find the value of $$x$$.

A
$$10$$

B
$$15$$

C
$$20$$

D
$$25$$

Solution

$$\Rightarrow$$ The given numbers are $$\dfrac{x}{3},\dfrac{x}{2},\dfrac{x}{4},\dfrac{2x}{9},x$$
Arranging given numbers in ascending order,
$$\dfrac{2x}{9},\dfrac{x}{4},\dfrac{x}{3},\dfrac{x}{2},x$$
$$\Rightarrow$$ Now, here we can see $$\dfrac{x}{3}$$ is middle number.
According to the question,
$$\therefore$$ $$\dfrac{x}{3}=5$$
$$\therefore$$ $$x=15$$
Question 6
1 / -0

Find the median of $$15\dfrac {2}{3}, 15.03, 15, 15\dfrac {1}{3}$$ and $$15.3$$

A
$$15\dfrac {1}{3}$$

B
$$15.3$$

C
$$15\dfrac {2}{3}$$

D
$$15.03$$

Solution

Arranging in scaling ascending order, we get observations as
$$15,15.03,15.30,15.50,15\cfrac{2}{3}$$
so, median $$=15.30$$

Question 7

1 / -0

Directions For Questions

Study the following table and answer the questions based on it.
Number of Candidates Appeared, Qualified and Selected in a Competitive Examination from Five States Delhi, H.P, U.P, Punjab and Haryana Over the Years $$1994$$ to $$1998$$

		Delhi			H.P.			U.P			Punjab			Haryana
Year	App	Qual	Sel	App	Qual	Sel	App	Qual	Sel	App	Qual	Sel	App	Qual	Sel
$$1997$$	$$8000$$	$$850$$	$$94$$	$$7800$$	$$810$$	$$82$$	$$7500$$	$$720$$	$$79$$	$$8200$$	$$680$$	$$85$$	$$6400$$	$$700$$	$$75$$
$$1998$$	$$4800$$	$$500$$	$$48$$	$$7500$$	$$800$$	$$65$$	$$5600$$	$$620$$	$$85$$	$$6800$$	$$600$$	$$70$$	$$7100$$	$$650$$	$$75$$
$$1999$$	$$7500$$	$$640$$	$$82$$	$$7400$$	$$560$$	$$70$$	$$4800$$	$$400$$	$$48$$	$$6500$$	$$525$$	$$65$$	$$5200$$	$$350$$	$$55$$
$$2000$$	$$9500$$	$$850$$	$$90$$	$$8800$$	$$920$$	$$86$$	$$7000$$	$$650$$	$$70$$	$$7800$$	$$720$$	$$84$$	$$6400$$	$$540$$	$$60$$
$$2001$$	$$9000$$	$$800$$	$$70$$	$$7200$$	$$850$$	$$75$$	$$8500$$	$$950$$	$$80$$	$$5700$$	$$485$$	$$60$$	$$4500$$	$$600$$	$$75$$

...view full instructions

For which state the average number of candidates selected over the years is the maximum?

Delhi

H.P

U.P

Punjab

Solution

The average number of candidates selected over the given period for various states are:

For Delhi $$= \dfrac {94 + 48 + 82 + 90 + 70}{5} $$

$$= \dfrac {385}{5} = 76.8$$

For U.P. $$= \dfrac {78 + 85 + 48 + 70 + 80}{5} $$

$$= \dfrac {361}{5} = 72.2$$

For Punjab $$= \dfrac {85 + 70 + 65 + 84 + 60}{5} $$

$$= \dfrac {364}{5} = 72.8$$

For Haryana $$= \dfrac {75 + 75 + 55 + 60 + 75}{5} $$

$$= \dfrac {340}{5} = 68$$

The average is maximum for Delhi.

Question 8
1 / -0

What is the median of the values $$11, 7, 6, 9, 12, 15, 19$$?

A
$$9$$

B
$$11$$

C
$$12$$

D
$$15$$

Solution

To find median, we write all the scores from smallest to largest and then choose the score in the middle.
Ordered data : $$6,7,9,11,12,15,19$$
Therefore, median is $$11$$.

Question 9

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Directions For Questions

Study the following table and answer the questions based on it.
Expenditure of a Company (in Lakh Rupees) per Annum Over the given years.

Year	Salary	Fuel and Transport	Bonus	Interest on Loans	Taxes
$$1998$$	$$288$$	$$98$$	$$3.00$$	$$23.4$$	$$83$$
$$1999$$	$$342$$	$$112$$	$$2.52$$	$$32.5$$	$$108$$
$$2000$$	$$324$$	$$101$$	$$3.84$$	$$41.6$$	$$74$$
$$2001$$	$$336$$	$$133$$	$$3.68$$	$$36.4$$	$$88$$
$$2002$$	$$420$$	$$142$$	$$3.96$$	$$49.4$$	$$98$$

...view full instructions

What is the average amount of interest per year which the company had to pay during this period?

$$Rs. 32.43$$ lakhs

$$Rs. 33.72$$ lakhs

$$Rs. 34.18$$ lakhs

$$Rs. 36.66$$ lakhs

Solution

Average amount of interest paid by the Company during the given period
$$= Rs. \left [\dfrac {23.4 + 32.5 + 41.6 + 36.4 + 49.4}{5}\right ]lakhs$$
$$= Rs. \left [\dfrac {183.3}{5}\right ]lakhs$$
$$= Rs. 36.66\ lakhs$$

Question 10
1 / -0

Mean weight of a class is $$50Kg$$. Mean weight of boys and girls are $$52Kg$$ and $$42Kg$$ respectively.
Find the ratio of number of boys and girls in the school.

A
$$5:1$$

B
$$4:1$$

C
$$3:2$$

D
$$3:5$$

Solution

$$u_g=42$$, $$u_b=52$$, $$u=50$$
Assume the total strength of class to be 100.
So, $$n_b=n$$ and $$n_g=100-n$$
$$u=\dfrac{u_g \times n_g + u_b \times n_b}{n_g+n_b}$$
$$\therefore 50=\dfrac{42 \times (100-n)+ 52 \times n}{100}$$
$$\therefore 50=\dfrac{52n+4200-42n}{100}$$
$$\therefore 800=10n$$
Thus, $$n=80$$ and $$100-n=20$$
Hence the ratio : $$(4:1)$$
Ans-Option B.

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Measures of Central Tendency Test - 19

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Measures of Central Tendency Test - 19

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Question 1

0.4

0.5

0.6

0.8

Solution

Question 2

$$31$$

$$32$$

$$33$$

$$34$$

Solution

Question 3

median and mean

mean and mode

median and mode

mode and range

Solution

Question 4

$$9$$

$$8$$

$$7.5$$

$$7$$

Solution

Question 5

$$10$$

$$15$$

$$20$$

$$25$$

Solution

Question 6

$$15\dfrac {1}{3}$$

$$15.3$$

$$15\dfrac {2}{3}$$

$$15.03$$

Solution

Question 7

Delhi

H.P

U.P

Punjab

Solution

Question 8

$$9$$

$$11$$

$$12$$

$$15$$

Solution

Question 9

$$Rs. 32.43$$ lakhs

$$Rs. 33.72$$ lakhs

$$Rs. 34.18$$ lakhs

$$Rs. 36.66$$ lakhs

Solution

Question 10

$$5:1$$

$$4:1$$

$$3:2$$

$$3:5$$

Solution

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