Statistics Test

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Statistics Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

The median of the data 30, 25, 27, 29, 35, 38, 28 is _______.

A
28.5

B
29.5

C
28

D
29

Solution

Arrange the given data in ascending order.
We have, $$25, 27, 28, 29, 30, 35, 38$$
Median is $$29$$
Question 2
1 / -0

The difference between the maximum and the minimum observations in the data is

A
class interval

B
frequency

C
cumulative frequency

D
range

Solution

The difference between maximum and the minimum observation in the data is range.
For example, suppose an experiment involves finding out the weight of lab rats and the values in grams are 320, 367, 423, 471 and 480. In this case, the range is simply computed as 480-320 = 160 grams.
Question 3
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.
Question 4
1 / -0

Which of the following is correct for the given data $$55, 38, 69, 24, 89$$?

A
$$median=mode$$

B
$$mean=mode$$

C
$$mean=median$$

D
None of these

Solution

$$24, 38, 55, 69, 89$$

$$Mean = \dfrac{sum\ of \ all \ the\ terms}{total\ number\ of\ terms} $$

$$Mean = \dfrac{24+38+55+69+89}{5}$$

$$Mean=\dfrac {275}{5}=55$$
if n (total number of terms ) is odd then the middle term when the data is arranged in ascending order is median
Arranging the data in ascending order, we get :
$$24, 38, 55, 69, 89$$
$$Median=55$$
Therefore, for the given data,$$Mean=median$$
Question 5
1 / -0

The mean deviation of $$\displaystyle a^{3}+b^{3}\: and\: a^{3}-b^{3}$$ (when (a & b > 0) is

A
$$\displaystyle a^{3}$$

B
$$\displaystyle b^{3}$$

C
$$\displaystyle 2a^{3}$$

D
$$\displaystyle 2b^{3}$$

Solution

Mean deviation is the mean of the absolute differences between each value in the data set and the mean of the data set.

Mean is the sum of the values in the data set divided by the total number of values.

Mean of $$a^3+b^3$$ and $$a^3-b^3$$ is $$\dfrac{a^3+b^3+a^3-b^3}{2}$$

$$\implies mean =\dfrac{2a^3}{2}=a^3$$

Absolute difference between the data values and mean is $$|a^3+b^3-a^3|$$ and $$|a^3-b^3-a^3|$$

Therefore, the data values after finding the absolute differences are $$b^3$$ and $$b^3$$

Therefore, the mean deviation is $$\dfrac{b^3+b^3}{2}=b^3$$
Question 6
1 / -0

Consider the following groups A and B
A : 3, 4, 5,...... upto n terms
B : 15, 19, 23,..... upto n terms
If the mean deviations of groups A and B about their means are $$\displaystyle \alpha \: \: \: and\: \: \: \beta $$ respectively then

A
$$\displaystyle \beta=5\alpha $$

B
$$\displaystyle \beta=4\alpha+3 $$

C
$$\displaystyle \beta=4\alpha$$

D
None

Solution

$$A$$ $$B$$
$$3$$ $$15$$
$$4$$ $$19$$
$$5$$ $$23$$
$$.$$ $$.$$
$$.$$ $$.$$
$$.$$ $$.$$
$$n$$ $$n$$

$$\dfrac{\sum A}{n}=\alpha$$

$$\dfrac{\sum B}{n}=\beta$$
now for $$A$$
$$a=3$$
$$d=1$$
$$\sum A=\dfrac{n}{2}[2a+(n-1)d]$$

=$$\dfrac{n}{2}[6+(n-1)])$$

=$$\dfrac{n}{2}[5+n]$$

$$\alpha=\dfrac{\sum A}{n}=\dfrac{5+n}{2}$$

=>n=$$2\alpha-5$$
for $$B$$
$$a=15$$
$$d=4$$
$$\sum B=\dfrac{n}{2}[2a+(n-1)d]$$
=$$\dfrac{n}{2}[30+(n-1)4]$$
=$$\dfrac{n}{2}[26+4n]$$
$$\beta=\dfrac{\sum B}{n}=\dfrac{26+4n}{2}$$
$$\beta=\dfrac{26+4(2\alpha-5)}{2}$$
$$2\beta=6+8\alpha$$
$$\beta=4\alpha+3$$
Question 7
1 / -0

In a class test in English $$10$$ students scored $$75$$ marks $$12$$ students scored $$60$$ marks $$8$$ scored $$40$$ marks and $$3$$ scored $$30$$ marks the mode for their score is:

A
$$75$$

B
$$30$$

C
$$60$$

D
$$25$$

Solution

$$\textbf{Step -1: Defining mode.}$$
$$\text{In a given data set, the mode is the value that appears most often among every other values.}$$
$$\text{Given data set says, }$$
$$\text{10 Students scored 75 marks.}$$
$$\text{12 Students scored 60 marks.}$$
$$\text{8 Students scored 40 marks.}$$
$$\text{3 Students scored 30 marks.}$$
$$\textbf{Step -2: Finding the mode of their score.}$$
$$\text{The mode of their score will be the marks occurring most often.}$$
$$\therefore \text{12 Students scored 60 marks.}$$
$$\textbf{Hence, The mode of their score is 60. (Option C)}$$
Question 8
1 / -0

The mean of $$20$$ observations is $$15$$. One observation $$20$$ is deleted and two more observations are included to the data. If the mean of new set of observations is $$15$$, then find the sum of the two new observations included.

A
$$30$$

B
$$35$$

C
$$33$$

D
$$32$$

Solution

Mean $$ = \dfrac { \text{Sum of observations}}{\text{Total number of observations}} $$
Given, mean of $$ 20 $$ observations $$ = 15 $$.
Thus, sum of $$ 20 $$ observations $$ = 15 \times 20 = 300 $$
When $$ 20 $$ is deleted and two more observations are included in the data, sum of observation $$ = 300 - 20 + X + Y = 280 + X + Y $$.
And number of observations is $$ 20 - 1 + 2 = 21 $$
Given, Mean of wages of new set of observations $$ = 15 $$
$$ = \dfrac {280 + X + Y }{21} = 15 $$
$$ \Rightarrow 280 + X + Y= 15 \times 21 = 315 $$
$$ \Rightarrow X + Y = 315 - 280 = 35 $$
Question 9
1 / -0

The standard deviation of $$3x + 5, 3x + 7, 3x + 9, 3x + 11, 3x + 13, 3x + 15$$ and $$3x + 17$$ is ...........

A
$$2$$

B
$$3$$

C
$$4$$

D
$$1$$

Solution

Mean of the given numbers is $$\bar x =\dfrac{(3x+5)+(3x+7)+(3x+9)+(3x+11)+(3x+13)+(3x+15)+(3x+17)}{7}=\dfrac{21x+77}{7}=3x+11$$

total number of values $$n=7$$

Standard deviation is given by $$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n}}$$

$$\implies SD=\sqrt{\dfrac{(5-11)^2+(7-11)^2+(9-11)^2+(11-11)^2+(13-11)^2+(15-11)^2+(17-11)^2}{7}}$$

$$\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{7}}$$

$$\implies SD=\sqrt{\dfrac{112}{7}}$$

$$\implies SD=\sqrt{16}=4$$

Therefore the standard deviation for the given values is $$4$$
Question 10
1 / -0

For the given data, SD = 10, AM = 20, the coefficient
of variation is____

A
47

B
24

C
44

D
50

Solution

Coeffecient of variation $$ = \frac {SD}{AM} \times 100 = \frac {10}{20} \times 100 = 50 $$

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Re-Attempt Weekly Quiz Competition

Statistics Test - 29

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Statistics Test - 29

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SHARING IS CARING

Question 1

28.5

29.5

28

29

Solution

Question 2

class interval

frequency

cumulative frequency

range

Solution

Question 3

27

25

30

35

Solution

Question 4

$$median=mode$$

$$mean=mode$$

$$mean=median$$

None of these

Solution

Question 5

$$\displaystyle a^{3}$$

$$\displaystyle b^{3}$$

$$\displaystyle 2a^{3}$$

$$\displaystyle 2b^{3}$$

Solution

Question 6

$$\displaystyle \beta=5\alpha $$

$$\displaystyle \beta=4\alpha+3 $$

$$\displaystyle \beta=4\alpha$$

None

Solution

Question 7

$$75$$

$$30$$

$$60$$

$$25$$

Solution

Question 8

$$30$$

$$35$$

$$33$$

$$32$$

Solution

Question 9

$$2$$

$$3$$

$$4$$

$$1$$

Solution

Question 10

47

24

44

50

Solution

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