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

$$25$$

$$27$$

$$31$$

$$33$$

Solution

Question 3

$${ 10 }{ x }$$

$${ 10 }$$

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

0

Question 4

$$\dfrac{124}{3}$$

$$\dfrac{148}{3}$$

$$\dfrac{248}{3}$$

$$\dfrac{124}{9}$$

Solution

Question 5

$$20$$

$$21$$

$$22.36$$

$$24.70$$

Solution

Question 6

Mean

Median

Mode

Origin

Solution

Question 7

5

10

20

19

Solution

Question 8

$$10$$

$$15$$

$$12$$

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

Solution

Question 9

6

7

3

9

Solution

Question 10

$$36$$

$$48$$

$$24$$

$$18$$

Solution

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