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

101010

151515

121212

212\cfrac{21}{2}221​

Solution

Question 3

282828

303030

353535

383838

Solution

Question 4

21

26

25

27

Solution

Question 5

xˉ+yˉ\bar{x}+\bar{y}xˉ+yˉ​

xˉ+yˉ2\frac{\bar{x}+\bar{y}}{2}2xˉ+yˉ​​

xˉ+yˉn\frac{\bar{x}+\bar{y}}{n}nxˉ+yˉ​​

xˉ+yˉ2n\frac{\bar{x}+\bar{y}}{2n}2nxˉ+yˉ​​

Solution

Question 6

46.546.546.5

49.549.549.5

53.553.553.5

56.556.556.5

Solution

Question 7

Mean

Median

Cannot be concluded

None of these

Solution

Question 8

111111

161616

171717

212121

Solution

Question 9

7.87.87.8

888

8.28.28.2

8.48.48.4

Solution

Question 10

420420420

425425425

435435435

460460460

Solution

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Question Analysis

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$10$

$15$

$12$

$\cfrac{21}{2}$

$28$

$30$

$35$

$38$

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

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

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

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

$46.5$

$49.5$

$53.5$

$56.5$

$11$

$16$

$17$

$21$

$7.8$

$8$

$8.2$

$8.4$

$420$

$425$

$435$

$460$