Data Handling Test

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Data Handling Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The mean of 96, 104, 121, 134, 142, 149, 153 and 161 is 132.5
If true then enter $$1$$ and if false then enter $$0$$

A
Can't determine

B
0

C
1

D
None of these

Solution

The given observations are: 96, 104, 121, 134, 142, 149, 153 and 161

Mean = $$\frac{Sum}{Total}$$

Mean = $$\frac{96 + 104 + 121 + 134 + 142 + 149 + 153 + 161}{8}$$

Mean = $$132.5$$
Question 2
1 / -0

If the mode of the following data $$10, 11, 12, 10, 15, 14, 15, 13, 12, x, 9, 7$$ is $$15$$, then the value of $$x$$ is:

A
$$10$$

B
$$15$$

C
$$12$$

D
$$\cfrac{21}{2}$$

Solution

The mode is the value which occurs the most often in a given set of data.
In the data set provided here, the mode is 15, hence 15 should occur the most number of times.
$$10, 11, 12, 10, 15, 14, 15, 13, 12, x, 9, 7$$
Excluding $$x$$ and arranging these numbers in ascending order,
$$7, 9, 10, 10, 11, 12, 12, 13, 14, 15, 15$$
Here 10, 12 and 15 are occurring twice.
Since 15 is the mode, it should occur more than twice.
Therefore, the value of $$x$$ should be 15 if the value of mode is 15.
Question 3
1 / -0

The mean of five numbers is $$30$$. If one number is excluded, their mean becomes $$28$$. The excluded number is:

A
$$28$$

B
$$30$$

C
$$35$$

D
$$38$$

Solution

The mean of $$5$$ numbers $$=30$$
The sum of $$5$$ numbers is $$=30\times5 = 150$$

After excluding a number say, $$x$$
Mean $$= 28$$
Sum of four numbers $$= 28\times4 = 112$$

$$x = 150 -112 = 38$$
Question 4
1 / -0

The mean of prime numbers between 20 and 30 is :

A
21

B
26

C
25

D
27

Solution

The prime numbers between $$20$$ and $$30$$ are $$23,29$$

Mean of the data set is the average of values in the data set.

Therefore the mean of the prime numbers is $$\dfrac{23+29}{2}=\dfrac{52}{2}=26$$
Question 5
1 / -0

Let $$\bar{x}$$ be the mean of $$x_1, x_2 , ... , x_n$$ and $$\bar{y}$$ the mean of $$y_1, y_2, ... , y_n$$. If $$\bar{z}$$ is the mean of $$x_1, x_2, ... , x_n$$, $$y_1, y_2, ... , y_n$$, then $$\bar{z}$$ is equal to

A
$$\bar{x}+\bar{y}$$

B
$$\frac{\bar{x}+\bar{y}}{2}$$

C
$$\frac{\bar{x}+\bar{y}}{n}$$

D
$$\frac{\bar{x}+\bar{y}}{2n}$$

Solution

$$\bar{x}$$ is the mean of $$x_1, x_2 , ... , x_n$$, then
$$\bar{x}$$ = $$\dfrac{x_1 + x_2 + ... + x_n}{n}$$

$$\bar{y}$$ is the mean of $$y_1, y_2, ... , y_n$$. then
$$\bar{y}$$ =  $$\dfrac{y_1 + y_2 + ... + y_n}{n}$$

$$\bar{z}$$ is the mean of $$x_1, x_2, ... , x_n$$, $$y_1, y_2, ... , y_n$$,
$$\bar{z}$$ =  $$\dfrac{x_1 + x_2 + ... + x_n+ y_1 +y_2 + ...+y_n}{2n}$$
$$\bar{z}$$ =  $$\dfrac{\bar{x} + \bar{y}}{2}$$
Question 6
1 / -0

There are $$50$$ numbers. Each number is subtracted from $$53$$ and the mean of the numbers so obtained is found to be $$3.5$$. The mean of the given numbers is:

A
$$46.5$$

B
$$49.5$$

C
$$53.5$$

D
$$56.5$$

Solution

Total numbers $$= 50$$
Mean of numbers after subtracting $$53$$ from each $$= 3.5$$
Sum of numbers after subtracting $$53$$ from each $$=3.5\times50 = 175$$
Sum of the original numbers $$=175 + 53\times50 = 2825$$

Mean of the original numbers $$=\dfrac{2825}{50} = 56.5$$
Question 7
1 / -0

In a diagnostic test in mathematics given to students, the following marks (out of $$100$$) are recorded:
$$46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98, 44$$
Which average will be a good representative of the above data?

A
Mean

B
Median

C
Cannot be concluded

D
None of these

Solution

Median will be a good representative of the data, because $$(i)$$each value occurs once, $$(ii)$$the data is influenced by extreme values. Hence, option $$B$$ is the correct answer.
Question 8
1 / -0

Find the mean of the data $$10, 15, 17, 19, 20$$ and $$21$$.

A
$$11$$

B
$$16$$

C
$$17$$

D
$$21$$

Solution

Data observations are: $$10, 15, 17, 19, 20\ and\ 21$$

mean = $$\dfrac {sum}{Number}$$

mean = $$\dfrac {10 + 15 + 17 + 19+ 20+ 21}{6}$$

mean = $$\dfrac {102}{6}$$ = $$17$$
Question 9
1 / -0

Find the mean of:
$$\displaystyle 3\frac{2}{3}, 6\frac{1}{3}, 7\frac{2}{3}, 10\frac{1}{3}$$ and $$11$$

A
$$7.8$$

B
$$8$$

C
$$8.2$$

D
$$8.4$$

Solution

We need to find mean of $$3\dfrac {2}{3}, 6\dfrac {1}{3}, 7\dfrac {2}{3}, 10\dfrac {1}{3},11$$
We can write all as
$$3\dfrac {2}{3}=\dfrac {11}{3}, 6\dfrac {1}{3}=\dfrac {19}{3}, 7\dfrac {2}{3}=\dfrac {23}{3}, 10\dfrac {1}{3}=\dfrac {31}{3}, 11=\dfrac {33}{3}$$

Therefore, mean $$=\dfrac {11+19+23+31+33}{3\times 5}$$

$$=\dfrac {117}{15}$$

$$=7.8$$
Question 10
1 / -0

Find the mean of the observations $$425, 430, 435, 440, 445, ........... 495.$$

A
$$420$$

B
$$425$$

C
$$435$$

D
$$460$$

Solution

No. of observations$$=15$$
Sum of observations$$=425+430+435+440+445+450+455+460+465+470+475+480+485$$
$$+490+495=6900$$

Mean$$=\dfrac { 6900 }{ 15 } =\quad 460
$$

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Data Handling Test - 19

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Data Handling Test - 19

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Question 1

Can't determine

0

1

None of these

Solution

Question 2

$$10$$

$$15$$

$$12$$

$$\cfrac{21}{2}$$

Solution

Question 3

$$28$$

$$30$$

$$35$$

$$38$$

Solution

Question 4

21

26

25

27

Solution

Question 5

$$\bar{x}+\bar{y}$$

$$\frac{\bar{x}+\bar{y}}{2}$$

$$\frac{\bar{x}+\bar{y}}{n}$$

$$\frac{\bar{x}+\bar{y}}{2n}$$

Solution

Question 6

$$46.5$$

$$49.5$$

$$53.5$$

$$56.5$$

Solution

Question 7

Mean

Median

Cannot be concluded

None of these

Solution

Question 8

$$11$$

$$16$$

$$17$$

$$21$$

Solution

Question 9

$$7.8$$

$$8$$

$$8.2$$

$$8.4$$

Solution

Question 10

$$420$$

$$425$$

$$435$$

$$460$$

Solution

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