Measures of Central Tendency Test

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Measures of Central Tendency Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is not central tendency?

A
Mode

B
Arithmetic mean

C
Median

D
None

Solution

Central tendency of any data is defined by mean or median or mode.

Hence Mean deviation is not a central tendency.
Question 2
1 / -0

If the mean of 1,7,5,3,4,4 is m and 3,2,4,2,3,3, p is m-1 and median is q, then the value of p and q will be-

A
$$p = 3, q = 4$$

B
$$p=4, q = 3$$

C
$$p = 4, q = 4$$

D
$$p = 4 , q = 2$$

Solution

$$\displaystyle \frac{1+7+5+3+4+4}{6}=m$$
$$\displaystyle m=\frac{24}{6}=4$$
and, $$\displaystyle \frac{3+2+4+2+3+3+p}{7}=m-1$$
$$\displaystyle \frac{17+p}{7}=4-1$$
17+p=21
p=4
Now, median of 3,2,4,2,3,3,p, is q. Then,
For n=7(odd)
Median$$\displaystyle =\frac{n+1}{2}$$th term
$$\displaystyle q=\left ( \frac{7+1}{2} \right )$$ th term
q = 4th term
q = 2
Hence, $$p= 4$$
$$q = 2$$
Question 3
1 / -0

The weighted arithmetic mean of the first n natural numbers whose weights are equal to the corresponding numbers is given by

A
$$\displaystyle \frac{1}{2}\left ( n+1 \right )$$

B
$$\displaystyle \frac{1}{2}\left ( n+2 \right )$$

C
$$\displaystyle \frac{1}{3}\left ( 2n+1 \right )$$

D
$$\displaystyle \frac{1}{3}n\left ( 2n+1 \right )$$

Solution

The first 'n' natural numbers are 1,2,3,......,n
and their corresponding weights are 1,2,3,...,n
$$\therefore Weighted\ Mean=\bar{x}_w=\dfrac{\sum_{i=1}^{n}w_ix_i}{\sum_{i=1}^{n}w_i}$$
$$\Rightarrow \dfrac{1\times 1+2\times 2+3\times 3+.....+n\times n}{1+2+3+.....+n}$$
$$\Rightarrow \dfrac{1^2+2^2+3^3+.......+n^2}{1+2+3+...+n}$$
As sum of square of n natural numbers is $$=\dfrac{n(n+1)(2n+1)}{6}$$
$$\therefore mean= \dfrac{\dfrac{1}{6}n(n+1)(2n+1)}{\dfrac{1}{2}n(n+1)}$$
$$\Rightarrow \dfrac{1}{3}(2n+1)$$
Question 4
1 / -0

Find the median of the following data
$$25, 21, 18, 27, 29, 16, 22, 15$$

A
$$21.5$$

B
$$27.5$$

C
$$21$$

D
$$16$$

Solution

Arranging the data in ascending order of magnitude
$$15, 16, 18, 21, 22, 25, 27, 29$$

Since there are $$8$$, i.e. an even number of observation

therefore, median = arithmetic mean of $$\left(\displaystyle\frac{8}{2}\right)^{th}$$ and $$\left(\displaystyle\frac{8}{2} + 1\right)^{th}$$ observation

Median = $$\displaystyle\frac{21+22}{2} = 21.5$$
Question 5
1 / -0

Which of the following is correct for the given data $$-1, 0, 1, 2, 3, 5, 5, 6, 8, 10, 11, ?$$

A
Mean=Mode=Median

B
Mean $$=5$$

C
Mean=Mode

D
Mode=Median

Solution

$$\Rightarrow$$ The given numbers are $$-1,0,1,2,3,5,5,6,8,10,11$$
$$\Rightarrow$$ $$Mean=\dfrac{-1+0+1+2+3+5+5+6+8+10+11}{11}=4.54$$
$$\Rightarrow$$ Here, middle number is $$6^{th}$$ number which is $$5$$.
$$\therefore$$ $$Median=5$$
$$\Rightarrow$$ In given numbers $$5$$ occurs two times.
$$\therefore$$ $$Mode=5$$
From above we can see that,
$$\therefore$$ $$Mode=Median$$
Question 6
1 / -0

The median of 3, 5, 6, x, 9,15 is 7. The value of x is

A
$$7$$

B
$$4$$

C
$$8$$

D
none of these

Solution

$$Median=\dfrac{6+x}{2}$$
$$\Rightarrow 7=\dfrac{6+x}{2}$$
$$\Rightarrow 14=6+x$$
$$\Rightarrow x=14-6$$
$$\Rightarrow x=8$$
Question 7
1 / -0

In an examination a candidate scores the following percentage of marks English-44; Hindi-58; Maths-74; Physics-61; Chemistry-62 If weights 2,4,4,5,3 respectively are allotted to these subjects then the candidate's weighted mean percentage is

A
$$61$$

B
$$61.5$$

C
$$62$$

D
$$62.5$$

Solution

Percentage marks in English, Hindi,Maths,Physics,Chemistry=44,58,74,61,62
Their weight=2,4,4,5,3
$$\therefore Weighted mean=\dfrac{2\times 44+4\times 58+4\times 74+61\times 5+62\times 3}{2+4+4+5+3}$$
$$\Rightarrow \dfrac{88+232+296+305+186}{18}=\dfrac{1107}{18}=61.5$$
Question 8
1 / -0

Mean of 10 values is 32.6. If another values is included the mean becomes 31. The included value is

A
$$16$$

B
$$14$$

C
$$15$$

D
$$19$$

Solution

Included value $$\displaystyle =31\times 11-32.6\times 10=15$$
Question 9
1 / -0

$$\displaystyle{x}=a+\frac{\Sigma{fd}}{N}$$ is the formula of

A
Median

B
Mode

C
Arithmetic mean

D
Mean deviation

Solution

The given formula is to determine Arithmetic mean using assumed mean method, where $$ a $$ is the assumed mean.
Question 10
1 / -0

A candidate obtained the following percentage of marks in an examination English 60, Maths 90, Physics 75, Chemistry 66. If weight 2,4,3,3 are allotted to these subjects respectively, then the weight mean is given by

A
$$\displaystyle \frac{60+90+75+66}{2+4+3+3}$$

B
$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times3 }{2+4+3+3}$$

C
$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times 3}{4}$$

D
$$\displaystyle \frac{60+90+75+66}{4}$$

Solution

Percentage marks in the examination-
English=60
Maths=90
Physics=75
Chemistry=66
Weight allotted to each subjects are-
English=2,Maths=4,English=3 and chemistry=66
Mean=$$\dfrac{sum\ of\ all\ observation\ value}{Sum\ of\ all\ observation}$$
$$\therefore$$ Weight mean=$$\dfrac{60\times 2+90\times 4+75\times 3+66\times 3}{2+4+3+3}$$

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Measures of Central Tendency Test - 14

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Measures of Central Tendency Test - 14

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Question 1

Mode

Arithmetic mean

Median

None

Solution

Question 2

$$p = 3, q = 4$$

$$p=4, q = 3$$

$$p = 4, q = 4$$

$$p = 4 , q = 2$$

Solution

Question 3

$$\displaystyle \frac{1}{2}\left ( n+1 \right )$$

$$\displaystyle \frac{1}{2}\left ( n+2 \right )$$

$$\displaystyle \frac{1}{3}\left ( 2n+1 \right )$$

$$\displaystyle \frac{1}{3}n\left ( 2n+1 \right )$$

Solution

Question 4

$$21.5$$

$$27.5$$

$$21$$

$$16$$

Solution

Question 5

Mean=Mode=Median

Mean $$=5$$

Mean=Mode

Mode=Median

Solution

Question 6

$$7$$

$$4$$

$$8$$

none of these

Solution

Question 7

$$61$$

$$61.5$$

$$62$$

$$62.5$$

Solution

Question 8

$$16$$

$$14$$

$$15$$

$$19$$

Solution

Question 9

Median

Mode

Arithmetic mean

Mean deviation

Solution

Question 10

$$\displaystyle \frac{60+90+75+66}{2+4+3+3}$$

$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times3 }{2+4+3+3}$$

$$\displaystyle \frac{60\times 2+90\times 4+75\times 3+66\times 3}{4}$$

$$\displaystyle \frac{60+90+75+66}{4}$$

Solution

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