Measures of Central Tendency Test

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Measures of Central Tendency Test - 24

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Question 1
1 / -0

A football player scored the following number of goals in the $$10$$ matches:
$$1, 3, 2, 5, 8, 6, 1, 4, 7, 9$$
Since the number of matches is $$10$$ (an even number), therefore, the median
$$=\dfrac{5^{th} \text{observation} +6^{th} \text{observation}}{2}$$
$$=\dfrac{8+6}{2}=7$$
Is it the correct answer and why?

A
Correct

B
Incorrect

C
Ambiguous

D
Data insufficient

Solution

Data have first to be arranged in ascending (or descending) order before finding the median.
Hence, option $$B$$ is the correct answer.
Question 2
1 / -0

If the mean of variate X in distribution 2.6, then the value of k is
Variate X 1 2 3 4 5
Frequency of X 4 5 k 1 2

A
$$8$$

B
$$10$$

C
$$12$$

D
$$18$$

Solution

$$\dfrac{1\times 4+2 \times 5+3\times k+4\times 1+5\times 2}{4+5+k+1+2}=2.6$$
$$\Rightarrow 0.4k=3.2 \Rightarrow k=8$$
Question 3
1 / -0

The points scored by a basket ball team in a series of matches are as follows:
$$17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 48, 10, 8, 7, 10, 28$$
Find the median and mode for the data.

A
Median $$= 12$$, mode $$= 10$$

B
Median $$= 11$$, mode $$= 10$$

C
Median $$= 12$$, mode $$= 18$$

D
Median $$= 10$$, mode $$= 18$$

Solution

Arrange in ascending order $$=2,5,7,7,8,10,10,10,14,17,18,24,25,27,28,48$$
Total $$16$$ elements
So median $$=\left(\dfrac{(8^{th}+9^{th})\: \: \text{element}}{2}\right)$$
$$=\dfrac{10+14}{2}$$
$$=12$$
mode $$ = 10$$ $$(3$$ times repeating$$)$$
Question 4
1 / -0

A total of $$25$$ patients admitted to a hospital are tested for levels of blood sugar $$(mg/dl)$$ and the results obtained were as follows:
$$87$$ $$71$$ $$83$$ $$67$$ $$85$$
$$77$$ $$69$$ $$76$$ $$65$$ $$85$$
$$85$$ $$54$$ $$70$$ $$68$$ $$80$$
$$73$$ $$78$$ $$68$$ $$85$$ $$73$$
$$81$$ $$78$$ $$81$$ $$77$$ $$75$$
Find mean, median and mode $$(mg/dl)$$ of the above data.

A
Mean $$= 75.64$$, Median $$= 77$$, Mode $$= 85$$

B
Mean $$= 75.64$$, Median $$= 76$$, Mode $$= 81$$

C
Mean $$= 75.64$$, Median $$= 77$$, Mode $$= 81$$

D
Mean $$= 73.64$$, Median $$= 76$$, Mode $$= 85$$

Solution

Arrange them in ascending order,
$$54,65,67,68,68,69,70,71,73,73,75,76,77,77,78,78,80,81,81,83,85,85,85,85,87$$
so total no. $$=25$$
sum of total no. $$=1891$$
$$\therefore$$ mean $$=\dfrac{1891}{25}=75.64$$
median $$=\left(\dfrac{25+1}{2}\right)^{th}$$ element
$$= 13^{th}$$ element
$$=77$$
mode $$=$$ highest frequency element $$=85$$
Question 5
1 / -0

The mean marks (out of $$100$$) of boys and girls in an examination are $$70$$ and $$73$$, respectively. If the mean marks of all the students in that examination is $$71$$, find the ratio of the number of boys to the number of girls.

A
$$1:1$$

B
$$2:1$$

C
$$1:2$$

D
$$2:3$$

Solution

Let the total number of boys be $$x$$ and the total number of girls br $$y$$.
Let the sum of the marks of boys be $$a$$ and the sum of the marks of girls be $$b$$.
Mean marks of boys are $$=70$$
Mean marks of girls are $$=73$$
Therefore,
$$\Rightarrow \dfrac{a}{x}=70$$
$$\Rightarrow a=70x$$
and
$$\Rightarrow \dfrac{b}{y}=73$$
$$\Rightarrow b=73y$$
Now, mean marks of all the students is $$71$$
Therfore,
$$\Rightarrow \dfrac{sum\ of\ the\ marks\ of\ all\ the\ students}{x+y}=71$$
$$\Rightarrow sum\ of\ the\ marks\ of\ all\ the\ students=a+b$$
$$\Rightarrow a+b=70x+73y$$
Therefore,
$$\Rightarrow \dfrac{70x+73y}{x+y}=71$$
$$\Rightarrow 70x+73y=71x+71y$$
$$\Rightarrow2y=x$$
$$\Rightarrow \dfrac{x}{y}=\dfrac{2}{1}$$
$$Hence\ the\ ratio\ is\ 2:1$$
Question 6
1 / -0

Mean of $$50$$ observations was found to be $$80.4$$. But later on, it was discovered that $$96$$ was misread as$$ 69$$ at one place. Find the correct mean.

A
$$75.9$$

B
$$8.4$$

C
$$79$$

D
$$80.94$$

Solution

$$\dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 } }{ 50 } =80.4$$
Say $${ x }_{ 10 }$$ (incorrect value) $$=69$$
$${ x' }_{ 10 }$$ (correct value) $$=96={ x }_{ 10 }+27$$
$${ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 }=4020$$
Add $$27$$ on both sides
$${ x }_{ 1 }+{ x }_{ 2 }+...+{ x }_{ 50 }+27=4020+27=4047$$
$${ x }_{ 1 }+{ x }_{ 2 }+...+{ x' }_{ 10 }+...+{ x }_{ 50 }=4047$$
Actual mean $$=\dfrac { { x }_{ 1 }+{ x }_{ 2 }+...+{ x' }_{ 10 }+...+{ x }_{ 50 } }{ 50 } $$
Therefore, $$\dfrac { 4047 }{ 50 } =80.94$$
Question 7
1 / -0

The median of the data
$$78, 56, 22, 34, 45, 54, 39, 68, 54, 84$$ is

A
$$45$$

B
$$49.5$$

C
$$54$$

D
$$56$$

Solution

Terms are: $$78, 56, 22, 34, 45, 54, 39, 68, 54, 84.$$
Arranging the terms in ascending order: $$22,34,39,45,54,54,56,68,78,84$$.

Since the number of terms are even, the median will be the mean of $$5^{th}$$ and $$6^{th}$$ terms:
Median $$= \dfrac{54+54}{2} = 54$$
Question 8
1 / -0

The mean of $$25$$ observations is $$36$$. Out of these observations if the mean of first $$13$$ observations is $$32$$ and that of the last $$13$$ observations is $$40$$, the $$13$$th observation is :

A
$$23$$

B
$$36$$

C
$$38$$

D
$$40$$
Solution
Mean of $$25$$ observations =$$ 36$$
Sum of $$25$$ observations = $$36\times25 = 900$$
Mean of first $$13$$ observations= $$32$$
Sum of first $$13$$ observations = $$32\times13 = 416$$
Mean of Last $$13$$ observation=$$40$$
Sum of Last 13 observations = $$40\times13$$ = $$520$$

Sum of first$$13$$ observation+ Sum of last$$13$$ obs $$-$$ $$13$$th observation$$=$$ Total sum of $$25$$ observations
$$416 +520 - 13th\ observation = 900$$
- $$13$$ th observation $$ =36$$
Question 9
1 / -0

If the mean of the observations:
$$x, x + 3, x + 5, x + 7, x + 10$$ is $$9$$, the mean of the last three observations is

A
$$10\dfrac{1}{3}$$

B
$$10\dfrac{2}{3}$$

C
$$11\dfrac{1}{3}$$

D
$$11\dfrac{2}{3}$$

Solution

Mean of $$x, x + 3, x + 5, x + 7, x + 10 = 9$$
Sum of the terms $$= 5x + 25$$
Mean = $$\dfrac{5x+25}{5} =9$$
$$x +5 =9$$
$$x =4$$
Last three observations are $$9,11,14$$
Mean $$= \dfrac{9+11+14}{3}= 11\dfrac{1}{3}$$
Question 10
1 / -0

The mean of $$100$$ observations is $$50$$. If one of the observations which was $$50$$ is replaced by$$ 150$$, the resulting mean will be :

A
$$50.5$$

B
$$51$$

C
$$51.5$$

D
$$52$$

Solution

Mean of $$100$$ terms = $$50$$
Sum of $$100$$ terms = $$50\times 100$$ = 5000
After replacing a term by $$150$$, sum = $$5000 - 50 +150 = 5100$$
Mean = $$\dfrac{5100}{100}= 51$$

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Variate X	1	2	3	4	5
Frequency of X	4	5	k	1	2

Measures of Central Tendency Test - 24

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Measures of Central Tendency Test - 24

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Question 1

Correct

Incorrect

Ambiguous

Data insufficient

Solution

Question 2

$$8$$

$$10$$

$$12$$

$$18$$

Solution

Question 3

Median $$= 12$$, mode $$= 10$$

Median $$= 11$$, mode $$= 10$$

Median $$= 12$$, mode $$= 18$$

Median $$= 10$$, mode $$= 18$$

Solution

Question 4

Mean $$= 75.64$$, Median $$= 77$$, Mode $$= 85$$

Mean $$= 75.64$$, Median $$= 76$$, Mode $$= 81$$

Mean $$= 75.64$$, Median $$= 77$$, Mode $$= 81$$

Mean $$= 73.64$$, Median $$= 76$$, Mode $$= 85$$

Solution

Question 5

$$1:1$$

$$2:1$$

$$1:2$$

$$2:3$$

Solution

Question 6

$$75.9$$

$$8.4$$

$$79$$

$$80.94$$

Solution

Question 7

$$45$$

$$49.5$$

$$54$$

$$56$$

Solution

Question 8

$$23$$

$$36$$

$$38$$

$$40$$

Solution

Question 9

$$10\dfrac{1}{3}$$

$$10\dfrac{2}{3}$$

$$11\dfrac{1}{3}$$

$$11\dfrac{2}{3}$$

Solution

Question 10

$$50.5$$

$$51$$

$$51.5$$

$$52$$

Solution

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