Measures of Central Tendency Test

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Measures of Central Tendency Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

Find the upper quartile for the data: $$9,11,15,19,17,13,7.$$

A
$$17$$

B
$$16$$

C
$$12$$

D
None of these

Solution

Upper quartile is the median of the upper half of the data set.
Given data set is $$9,11,15,19,17,13,7$$
Arranging the data set in the ascending order we get $$7,9,11,13,15,17,19$$
Number of values in the data set is $$n=7$$

Upper quartile is given by $$Q_3=\dfrac 34(n+1)=\dfrac 34(7+1)=\dfrac 34(8)=6^{th} value$$

Therefore, the upper quartile is $$17$$
Question 2
1 / -0

Find the mode of the following data:
$$25, 16, 19, 48, 19, 20, 34, 15, 19, 20, 21, 24, 19, 16, 22, 16, 18, 20, 16, 19$$

A
$$19$$

B
$$18$$

C
$$17$$

D
$$25$$

Solution

Frequency table for the given data as given below
Value $$\displaystyle f_{i}$$: 15 16 18 19 20 21 22 24 25 34 48
Frequency$$\displaystyle f_{i}$$: 1 4 1 5 3 1 1 1 1 1 1
The observation with the maximum frequency is the mode of the series.
Here, 19 has the maximum frequency of 5.
So, Mode = 19
Question 3
1 / -0

Which of the following does not change for the observation $$23, 50, 27, 2x, 48, 59, 72, 89, 5x, 100, 120$$ when $$x$$ lies between $$15$$ and $$20$$?

A
Arithmetic mean

B
Range

C
Median

D
Quartile deviation

Solution

$$23, 50, 27, 2x, 48, 59, 72, 89, 5x, 100, 120$$

Now, when x lies in $$15$$ and $$20$$, the numbers $$2x$$ and $$5x$$ may range between $$30, 75$$ and $$40, 100$$ respectively.
Thus, there will be no effect on the minimum value and maximum value. Hence, the range of the series will not be affected.

Question 4

1 / -0

Mean of the following frequency distribution is

Class	0-5	5-10	10-15	15-20	20-25
Frequency	2	4	8	6	10

$$15.5$$

$$16.5$$

$$14.5$$

$$13.5$$

Solution

Class	Frequency(f)	Class Mark (x)	fx
0-5	2	2.5	5
5-10	4	7.5	30
10-15	8	12.5	100
15-20	6	17.5	105
20-25	10	22.5	225
Total	30		465

Mean $$ = \cfrac { \sum { fx } }{ \sum { f }} = \cfrac {465}{30} = 15.5 $$

Question 5

1 / -0

The mean of the following distribution is

Class	$$0-5$$	$$5-10$$	$$10-15$$	$$15-20$$	$$20-25$$	$$25-30$$
Frequency	$$4$$	$$5$$	$$7$$	$$12$$	$$7$$	$$5$$

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Answer:-

$$x_i=\cfrac{\text{upper limt + lower limit}}{2}$$

Class	Frequency	$$x_i$$	$$x_i.f_i$$
$$0-5$$	$$4$$	$$2.5$$	$$6$$
$$5-10$$	$$5$$	$$7.5$$	$$37.5$$
$$10-15$$	$$7$$	$$12.5$$	$$87.5$$
$$15-20$$	$$12$$	$$17.5$$	$$210$$
$$20-25$$	$$7$$	$$22.5$$	$$157.5$$
$$25-30$$	$$5$$	$$27.5 $$	$$137.5$$
	$$\Sigma f_i=40$$		$$\Sigma f_i.x_i=640$$

Mean =$$ \cfrac { \Sigma f_{ i }.x_{ i } }{ \Sigma f_{ i } } =\cfrac { 640 }{ 40 } =16$$

Question 6
1 / -0

The marks obtained by $$19$$ students of a class are $$27,36,22,31,25,26,33,24,37,32,29,28,36,35,27,26,32,35$$ and $$28$$. Find the lower quartile.

A
$$26$$

B
$$34$$

C
$$24$$

D
None of the above

Solution

By Arrange the series in increasing order. we get
$$22,24,25,26,26,27,27,28,28,29,31,32,32,33,35,35,36,36,37$$

Since, number of terms $$N=19$$
Lower quartile $$Q_1=\cfrac {N+1}{4}=\cfrac {19+1}{4}=5^{th} term$$
$$\therefore Q_1=26$$
Question 7
1 / -0

Find the median of the following data
$$10,18,15,11,9,16,8$$

A
$$11$$

B
$$18$$

C
$$9$$

D
$$16$$

Solution

Arranging the data in increasing order of magnitude
$$8, 9, 10, 11, 15, 16, 18$$

Since there are $$7$$, i.e. an odd number of observation,

therefore, median = value of $$\left(\displaystyle\frac{7+1}{2}\right)^{th}$$ observation = value of $$4^{th}$$ observation = $$11$$
Question 8
1 / -0

Median of $$4,5,10,7,1,14,9$$ and $$15$$ will be:

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

Answer:-
Arranging the data in sequence
$$1, 4, 5, 7, 9, 10, 14, 15$$.
Median = $$\cfrac{N+1}{2}$$
$$N$$ is no. of data given
Here $$N=8$$
$$\Rightarrow\; \cfrac{8+1}{2}=4.5^{th}\;term$$
$$\cfrac{ 4\;^{th}term + \;5^{th}term}{2} = \cfrac{\left(7+9\right)}{2}=\cfrac{16}{2}=8$$
Question 9
1 / -0

On $$13$$ consecutive days the number of person booked for violating speed limit of $$40$$ km/hr. were as follows $$59, 52, 58, 61, 68, 57, 62, 50, 55, 62, 53, 54, 51$$.
The median number of speed violations per day is

A
$$61$$

B
$$52$$

C
$$55$$

D
$$57$$

Solution

$$\Rightarrow$$ In the above given data $$62$$ is repeated twice,
$$\Rightarrow$$ Now we write the data in ascending order.
$$\Rightarrow$$ $$50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 62, 62, 68$$
$$\Rightarrow$$ Thus, the measure of $$\dfrac{13+1}{2}=7th$$ number is median value.
$$\therefore$$ The median number of speed violation per day is $$57$$.
Question 10
1 / -0

The arithmetic mean of 5 numbers is 27. lf one of the number is excluded the mean of the remaining number is 25. Find the excluded number.

A
27

B
25

C
30

D
35

Solution

Sum of 5 numbers =27 $$\times $$ 5 =135
When one of the numbers is excluded.
Sum of remaining 4 numbers = 4 $$\times $$ 25 = 100
Excluded number = 135 -100 = 35.

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Value $$\displaystyle f_{i}$$:	15	16	18	19	20	21	22	24	25	34	48
Frequency$$\displaystyle f_{i}$$:	1	4	1	5	3	1	1	1	1	1	1

Measures of Central Tendency Test - 33

Result

Measures of Central Tendency Test - 33

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Question 1

$$17$$

$$16$$

$$12$$

None of these

Solution

Question 2

$$19$$

$$18$$

$$17$$

$$25$$

Solution

Question 3

Arithmetic mean

Range

Median

Quartile deviation

Solution

Question 4

$$15.5$$

$$16.5$$

$$14.5$$

$$13.5$$

Solution

Question 5

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Question 6

$$26$$

$$34$$

$$24$$

None of the above

Solution

Question 7

$$11$$

$$18$$

$$9$$

$$16$$

Solution

Question 8

$$6$$

$$7$$

$$8$$

$$9$$

Solution

Question 9

$$61$$

$$52$$

$$55$$

$$57$$

Solution

Question 10

27

25

30

35

Solution

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