Measures of Central Tendency Test

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Measures of Central Tendency Test - 11

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Question 1
1 / -0

The weights of students of a certain class are given below.
$$39, 42, 47, 38, 42, 40, 42, 38, 43, 42, 38, 44, 46, 39, 42, 40, 43, 42, 41$$
Find the mode.

A
$$39$$

B
$$40$$

C
$$41$$

D
$$42$$

Solution

weight of students : $$39, 42, 47, 38, 42, 40, 42, 38, 43, 42, 38, 44, 46, 39, 42, 40, 43, 42, 41$$
Arranging them in ascending order: $$38, 38, 38, 39, 39, 40, 40, 41, 42,42,42,42,42,42, 43, 43, 44, 46, 47$$
$$42$$ occurs $$6$$ and maximum number of times in the data. Hence, $$42$$ is the modal weight.
Question 2
1 / -0

Find the mean and median of the data:
$$35, 48, 92, 76, 64, 52, 51, 63$$ and $$71$$.
If $$51$$ is replaced by $$66$$, what will be the new median?

A
$$65$$

B
$$64$$

C
$$66$$

D
$$67$$

Solution

Arranging data in ascending order- $$35, 48,51,52, 63, 64, 71,76, 92$$
Count of observations $$=9 $$
Mean $$=$$ Sum of all observations/count $$= \dfrac {552}{9}= 61.33$$

Median for odd number of data is the middle most observation
So median $$=$$ $$5^{th}$$ observation $$= 63$$

If $$51$$ is replaced by $$66$$, then the observations in ascending order will be- $$35, 48,52, 63, 64, 66,71,76, 92$$
Again Median for odd number of data is the middle most observation
So new median $$= 5^{th}$$ observation $$= 64$$
Question 3
1 / -0

The weights (in $$kg$$) of $$10$$ children are:
$$40,52,34,47,31,35,48,41,44,38$$. Find the median weight

A
$$35.9\:kg$$

B
$$38.4\:kg$$

C
$$40.5\:kg$$

D
$$42.5\:kg$$

Solution

Arranging the weights in ascending order, we have: $$31$$,$$34$$,$$35$$,$$38$$,$$40$$,$$41$$,$$44$$,$$47$$,$$48$$,$$52$$
Here, $$n = 10$$, which is even

$$\therefore$$ Median weight $$= \cfrac { 1 }{ 2 } \left\{ \begin{pmatrix} \cfrac { 10 }{ 2 } \end{pmatrix}^{th} \text{term}+\begin{pmatrix} \cfrac { 10 }{ 2 } +1 \end{pmatrix}^{th} \text{term} \right\} $$

$$=\cfrac { 1 }{ 2 } \left\{ 5^{th} \text{term}+6^{th} \text{term} \right\} =\cfrac { 1 }{ 2 } \left\{ 40+41 \right\}\: kg$$

$$= \cfrac { 81 }{ 2 }\:kg = 40.5\:kg$$

Hence, median weight $$= 40.5\:kg$$
Question 4
1 / -0

The median of the following data $$46$$, $$64$$, $$87$$, $$41$$, $$58$$, $$77$$, $$35$$, $$90$$, $$55$$, $$33$$, $$92$$ is

A
$$87$$

B
$$77$$

C
$$58$$

D
$$60.2$$

Solution

Given observations are: $$46,64,87,41,58,77,35,90,55,33,92$$
Arranging the observations in ascending order:
$$33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92$$
The series has $$11$$ observations, the median will the middle observation i.e. $$6^{th}$$ observation,
hence, the median $$= 58$$
Question 5
1 / -0

The following observations have been arranged in ascending order. If median of these observations is $$58$$, find the value of $$x$$.
$$24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90$$

A
$$58$$

B
$$57$$

C
$$55$$

D
$$59$$

Solution

Given series is: $$24,27,43,48, x - 1, x + 3, 68, 73, 80, 90$$
Since, the series has $$10$$ observations, the median will be the mean of $$5^{th}$$ and $$6^{th}$$ observation.
Median $$= \dfrac{x - 1 + x +3}{2}$$
Median $$= x +1$$
$$\Rightarrow 58 = x + 1$$
$$\Rightarrow x = 57$$
Question 6
1 / -0

The mean of first six multiples of $$4$$ is

A
$$13.5$$

B
$$14.5$$

C
$$14$$

D
$$16$$

Solution

$$\textbf{Step 1: Write all 6 multiples of 4}$$

$$\text{First six multiples of 4 are: }4, 8, 12, 16, 20, 24$$

$$\textbf{Step 2: Find Mean of all 6 observations}$$

$$\text{Mean }= \cfrac{4 + 8 + 12 + 16 + 20+ 24}{6}$$ $$\left[\because \boldsymbol{\textbf{Mean }= \cfrac{\textbf{Sum of observations}}{\textbf{Total observations}}}\right]$$

$$\text{Mean }= \cfrac{112}{6}$$

$$\text{Mean }= 14$$

$$\textbf{Hence, Option C is correct.}$$
Question 7
1 / -0

While computing mean of grouped data, we assume that the frequencies are:

A
evenly distributed over all the class

B
centred at the classmarks of the class

C
centred at the upper limits of the class

D
centred at the lower limits of the class

Solution

While computing mean of grouped data, we assume that the frequencies are centred at the class marks of the class.
Hence, option B is correct.

Question 8

1 / -0

Find the mean of the following data, by using the assumed mean method.

Class Interval	$$0-10$$	$$10-20$$	$$20-30$$	$$30-40$$	$$40-50$$
Frequency	$$7$$	$$8$$	$$12$$	$$13$$	$$10$$

$$10$$

$$30$$

$$27.2$$

None of these

Solution

Answer:-

Class interval	Frequency ($$f_i$$)	$$ x_i = \cfrac{\text{lower limit + upper limit}}{2} $$	$$ f_ix_i $$
$$0-10$$	$$ 7$$	$$5$$	$$35$$
$$10-20$$	$$8$$	$$15$$	$$120$$
$$20-30$$	$$12$$	$$25$$	$$300$$
$$30-40$$	$$13$$	$$35$$	$$455$$
$$40-50$$	$$10$$	$$45$$	$$450$$
	$$ \Sigma f_i = 50 $$		$$ \Sigma f_ix_i = 1360 $$

Mean = $$ \cfrac{\Sigma f_ix_i}{\Sigma f_i}= \cfrac{1360}{50} = 27.2 $$

Question 9
1 / -0

Find the median of:
$$233, 173, 189, 208, 194, 204, 194, 185, 200$$ and $$220.$$

A
$$199$$

B
$$200$$

C
$$197$$

D
$$198$$

Solution

Arrange the series in ascending order:
$$173, 185, 189, 194, 194, 200, 204, 208, 220, 233$$
The series has $$10$$ numbers, even numbers.
Hence, the median will be the mean of the two middle numbers
Median $$=$$ mean of $$5^{th}$$ and $$6^{th}$$ terms
Median $$= \dfrac{194 + 200}{2}$$
Median $$= 197$$
Question 10
1 / -0

To find mean, we use the formula

A
$$\displaystyle \sum_{i=1}^{n}f_ix_i$$

B
$$N\displaystyle \sum_{i=1}^{n}f_ix_i$$

C
$$\frac {1}{N}\displaystyle \sum_{i=1}^{n}f_ix_i$$

D
$$\displaystyle \sum_{i=1}^{n}\left (\frac {f_i(x_i)^2}{N}\right )$$

Solution

mean of the data = $$\frac{Sum \quad of \quad obeservations}{Number \quad of \quad observations}$$
If the data is a grouped data, frequency of each observation denoted by $$f_i$$ and observation by $$x_i$$, then
Mean = $$\frac {1}{N}\displaystyle \sum_{i=1}^{n}f_ix_i$$

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Re-Attempt Weekly Quiz Competition

Measures of Central Tendency Test - 11

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Measures of Central Tendency Test - 11

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SHARING IS CARING

Question 1

$$39$$

$$40$$

$$41$$

$$42$$

Solution

Question 2

$$65$$

$$64$$

$$66$$

$$67$$

Solution

Question 3

$$35.9\:kg$$

$$38.4\:kg$$

$$40.5\:kg$$

$$42.5\:kg$$

Solution

Question 4

$$87$$

$$77$$

$$58$$

$$60.2$$

Solution

Question 5

$$58$$

$$57$$

$$55$$

$$59$$

Solution

Question 6

$$13.5$$

$$14.5$$

$$14$$

$$16$$

Solution

Question 7

evenly distributed over all the class

centred at the classmarks of the class

centred at the upper limits of the class

centred at the lower limits of the class

Solution

Question 8

$$10$$

$$30$$

$$27.2$$

None of these

Solution

Question 9

$$199$$

$$200$$

$$197$$

$$198$$

Solution

Question 10

$$\displaystyle \sum_{i=1}^{n}f_ix_i$$

$$N\displaystyle \sum_{i=1}^{n}f_ix_i$$

$$\frac {1}{N}\displaystyle \sum_{i=1}^{n}f_ix_i$$

$$\displaystyle \sum_{i=1}^{n}\left (\frac {f_i(x_i)^2}{N}\right )$$

Solution

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