Statistics Test

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Statistics Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The mean weight of 9 students is 25 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is

A
25 kg

B
24 kg

C
26 kg

D
23 kg

Solution

Formula used:
$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$

Apply the above formula, we get the sum of weights of the 9 students = $25\times 9=225$ kg
If one more student is joined in the group, then total number of students is 10, and the mean isn 25.
The sum of weights of the 10 students is $25\times 10=250$ kg
The weight of the tenth student is $250-225=25$ kg
Question 2
1 / -0

The sum of $15$ observations is $434+x.$ If the mean of the data is $x,$ then find $x.$

A
$25$

B
$27$

C
$31$

D
$33$

Solution

Given, sum of all observations $=434+x$
Number of observations $=15$

and
$Mean=x$

$\Rightarrow \dfrac{434+x}{15}=x$

$\Rightarrow 434+x=15x$

$\Rightarrow x-15x=-434$

$\Rightarrow -14x=-434$

$\Rightarrow x=\dfrac{-434}{-14}$

$\Rightarrow x=31$

Hence, the value of $x$ is $31$ .
Question 3
1 / -0

$\sum _{ 1 }^{ 10 }{ x-\bar { x } }$ =..............

A
${ 10 }{ x }$

B
${ 10 }$

C
${ 9 }{ \bar { x } }$

D
0
Question 4
1 / -0

If $V$ is the variance and $M$ is the mean of first $15$ natural numbers, then what is $V+M^2$ equal to ?

A
$\dfrac{124}{3}$

B
$\dfrac{148}{3}$

C
$\dfrac{248}{3}$

D
$\dfrac{124}{9}$

Solution

first 15 natural numbers

$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$N=5$

mean,

$\mu =\dfrac{1}{N}\sum_{i=1}^{N}x_i$

$\mu =\dfrac{1}{15}(1+2+3+4+5+6+7+8+9+10+11+12+13+14+15)$

$=\dfrac{120}{5}=8$

variance,

$\sigma ^2=\dfrac{1}{N}\sum_{i=1}^{N}(x_i-\mu )^2$

$=\dfrac{1}{15}\sum_{i=1}^{15}(x_i-8 )^2$

$=\dfrac{1}{15}(280)$

Given,

$V+M^2$

$=\dfrac{280}{15}+8^2$

$=\dfrac{1240}{15}$

$=\dfrac{248}{3}$
Question 5
1 / -0

If the sum of squares of deviations of $15$ observations from their mean $20$ is $240$ , then what is the value of coefficient of variation (CV)?

A
$20$

B
$21$

C
$22.36$

D
$24.70$

Solution

$CV =\dfrac{sd}{mean} \times 100$

Standard deviation $sd = \sqrt {\dfrac{240}{15}}=\sqrt {16} = 4$

Mean $= 20$

$CV = \dfrac{4}{20} \times 100 = 0.2 \times 100 = 20$

Coefficient of variation $= 20$
Question 6
1 / -0

Mean deviation is lowest from

A
Mean

B
Median

C
Mode

D
Origin

Solution

(B) is correct
Question 7
1 / -0

The mean of first ten odd natural number is

A
5

B
10

C
20

D
19

Solution

Since, first ten odd natural numbers are $1, 3, 5, 7, 9, 11, 13, 15, 17, 19$

We know that
$\text{Mean}=\dfrac{\text{sum of observation}}{\text{number of observation}}$
$\therefore$ Mean $=\dfrac{(1 + 19 + 3 + 17 + 5 + 15 + 7 + 13 + 9 + 11) }{ 10} = \dfrac{100}{10} = 10$
Question 8
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 9
1 / -0

The mode of 9, 8, 7, 7, 6, 3, 7, 2, 1, 7, 9 is

A
6

B
7

C
3

D
9

Solution
Question 10
1 / -0

The highest score of a certain data exceeds in lowest score by $16$ and coefficient of range is $\cfrac{1}{3}$ . The sum of the highest score and the lowest score is

A
$36$

B
$48$

C
$24$

D
$18$

Solution

Let the highest score be $x_{m}$ and
the lowest score be $x_{0}$
Given that highest score exceeds lowest score by $16$
$\implies x_m=x_0+16\implies x_m-x_0=16$ ————(1)

Coefficient of range is given by $\dfrac{x_m-x_0}{x_m+x_0}$

Given that coefficient of range is $\dfrac 13$

$\implies \dfrac{x_m-x_0}{x_m+x_0}=\dfrac 13$ ———(2)

Substitute (1) in (2) we get

$\dfrac{16}{x_m+x_0}=\dfrac 13$

$\implies x_m+x_0=16\times3=48$

Therefore sum of the highest score and lowest score is $48$

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Re-Attempt Weekly Quiz Competition

Statistics Test - 23

Result

Statistics Test - 23

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Question 1

25 kg

24 kg

26 kg

23 kg

Solution

Question 2

252525

272727

313131

333333

Solution

Question 3

10x{ 10 }{ x }10x

10{ 10 }10

9xˉ{ 9 }{ \bar { x } }9xˉ

0

Question 4

1243\dfrac{124}{3}3124​

1483\dfrac{148}{3}3148​

2483\dfrac{248}{3}3248​

1249\dfrac{124}{9}9124​

Solution

Question 5

202020

212121

22.3622.3622.36

24.7024.7024.70

Solution

Question 6

Mean

Median

Mode

Origin

Solution

Question 7

5

10

20

19

Solution

Question 8

101010

151515

121212

212\cfrac{21}{2}221​

Solution

Question 9

6

7

3

9

Solution

Question 10

363636

484848

242424

181818

Solution

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Question Analysis

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$25$

$27$

$31$

$33$

${ 10 }{ x }$

${ 10 }$

${ 9 }{ \bar { x } }$

$\dfrac{124}{3}$

$\dfrac{148}{3}$

$\dfrac{248}{3}$

$\dfrac{124}{9}$

$20$

$21$

$22.36$

$24.70$

$10$

$15$

$12$

$\cfrac{21}{2}$

$36$

$48$

$24$

$18$