Measures of Central Tendency Test

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Measures of Central Tendency Test - 8

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Weekly Quiz Competition

Question 1
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Let the sum of the first three terms of an A.P. be $$39$$ and the sum of its last four terms be $$178$$. If the first term of this A.P. is $$10$$, then the median of the A.P. is:

A
$$29.5$$

B
$$28$$

C
$$31$$

D
$$26.5$$

Solution

Let the first three terms of the A.P. be $$a-d,a, a+d$$
Given $$a-d=10 ...(1)$$
Also given $$a-d+a+a+d=39$$
$$\Rightarrow 3a=39 ...(2)$$

Solving $$(1)$$ and $$(2)$$, we get
$$a=13, d=3$$
So, the A.P. is $$10,13,16,...$$

Now, let the last four terms be $$a_n, a_{n-1}, a_{n-2},a_{n-3}$$
$$a+(n-1)d+a+(n-2)d+a+(n-3)d+a+(n-4)d=178$$
$$\Rightarrow 4a+(4n-10)d=178$$
$$\Rightarrow n=14$$

So, total number of terms in the A.P. are $$14$$ (even).
Hence, median of the A.P. will be mean of $$T_7, T_8$$
$$T_7=10+6(3)=28$$
$$T_8=10+7(3)=31$$
Median $$=\cfrac{28+31}{2}=29.5$$
Question 2
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Median of the data: $$1, 9, 4, 3, 7, 6, 8, 8, 12, 15$$ is.

A
$$6.5$$

B
$$7.5$$

C
$$8$$

D
Any number between $$7$$ and $$8$$

Solution

First arrange the set of data in order(from least to highest)
$$1,3,4,6,7,8,8,9,12,15$$
As there are even number of items,
$$\therefore$$ Median $$=\dfrac{7+8}{2}=\dfrac{15}{2}=7.5$$
Question 3
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A set of numbers consists of three $$4$$'s, five $$5$$'s, six $$6$$'s, eight $$8$$'s and seven $$10$$'s. The mode of this set of numbers is

A
$$6$$

B
$$7$$

C
$$8$$

D
$$10$$

Solution

Since, $$8$$ occurs maximum number of times. It is the mode of the set.
Question 4
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A student obtained $$60, 75$$ and $$85$$ marks respectively in three monthly examinations in Mathematics and $$95$$ marks in the final examination. The three monthly examinations are of equal weightage whereas the final examination is weighted twice as much as a monthly examination. The mean marks of Mathematics are

A
$$80$$

B
$$81$$

C
$$82$$

D
$$85$$

Solution

We have
$$\quad x_1 = 60, x_2 = 75, x_3 = 85, x_4 = 95, w_1 = 1, w_2 = 1, w_3 = 1$$ and $$w_4 = 2$$
$$\therefore \quad \overline{X_w} = \displaystyle\frac{w_1x_1 + w_2x_2+ w_3x_3 + w_4x_4 }{w_1 + w_2 + w_3 + w_4}$$
$$\Rightarrow

\quad \overline{X_w} = \displaystyle\frac{1\times

60+1\times75+1\times85+2\times95}{1+1+1+2} = \displaystyle\frac{410}{5} =

82$$
Question 5
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Which of the following is not a measure of central tendency:

A
Mean

B
Median

C
Class interval

D
Mode

Solution

Mean, median & mode are most important measures of central tendency. Hence, option $$C$$ is correct.
Question 6
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The median of $$34,32,48,38,24,30,27,21,35$$ is

A
$$32$$

B
$$34$$

C
$$38$$

D
$$30$$

Solution

Arranging the data in ascending order of magnitude, we have
$$21,24,27,30,32,34,35,38,48$$
Since there are $$9$$, an odd number of items, Therefore the median is the
value of $$\left(\displaystyle\frac{9+1}{2}\right)^{th}$$ observation, i.e. $$32$$
Question 7
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Which one of the following is not central tendency ?

A
Mean deviation

B
Arithmetic mean

C
median

D
Mode

Solution

Mean deviation is deviation from the mean. It is not the central tendency.
Question 8
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Percentile divides the number of items into ______ equal parts.

A
$$10$$

B
$$100$$

C
$$20$$

D
$$50$$

Solution

A percentile (or a centile) is a measure used in statistics indicating the value below which a given percentage of observations in a group of observations fall.
For example, the 20th percentile is the value (or score) below which 20 percent of the observations may be found.
The term percentile and the related term percentile rank are often used in the reporting of scores from norm-referenced tests.
For example, if a score is in the 86th percentile, it is higher than 86% of the other scores.
100th percentile being the maximum, it therefore divides the items into 100 equal parts.
Question 9
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The median of $$10,\space 14,\space 11,\space 9,\space 8,\space 12,\space 6$$ is

A
$$14$$

B
$$11$$

C
$$10$$

D
$$12$$

Solution

Given numbers can be arranged in ascending order as,
$$6, 8, 9, 10,11, 12, 14 $$.
Number of term is $$7$$.
Thus median $$\displaystyle =\left(\frac{7+1}{2}\right)^{th} $$ term $$= 10$$
Question 10
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The mean of $$x, y, z$$ is $$y$$, then $$x+z =$$

A
$$y$$

B
$$2y$$

C
$$3y$$

D
$$zy$$

Solution

The question tells us that the mean of $$x, y$$ and $$z$$ is $$y$$.

i.e. $$ \dfrac {x+y+z}{3} =y $$

i.e. $$x+z=3y-y$$

$$\rightarrow x+z=2y$$

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Re-Attempt Weekly Quiz Competition

Measures of Central Tendency Test - 8

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Measures of Central Tendency Test - 8

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Question 1

$$29.5$$

$$28$$

$$31$$

$$26.5$$

Solution

Question 2

$$6.5$$

$$7.5$$

$$8$$

Any number between $$7$$ and $$8$$

Solution

Question 3

$$6$$

$$7$$

$$8$$

$$10$$

Solution

Question 4

$$80$$

$$81$$

$$82$$

$$85$$

Solution

Question 5

Mean

Median

Class interval

Mode

Solution

Question 6

$$32$$

$$34$$

$$38$$

$$30$$

Solution

Question 7

Mean deviation

Arithmetic mean

median

Mode

Solution

Question 8

$$10$$

$$100$$

$$20$$

$$50$$

Solution

Question 9

$$14$$

$$11$$

$$10$$

$$12$$

Solution

Question 10

$$y$$

$$2y$$

$$3y$$

$$zy$$

Solution

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