Statistics Test

Result

Statistics Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

Mean of variable series $$\bar{x} = 773$$ and mean deviation 64.4 , then coefficient of mean deviation is

A
$$0.065$$

B
$$12.003$$

C
$$0.083$$

D
$$0.073$$

Solution

$$\dfrac{Mean deviation}{Mean} = \dfrac{64.4}{773} = 0.083 $$
Thus (C) is correct
Question 2
1 / -0

Algebraic sum of deviation from mean is :

A
Negative

B
Positve

C
Different in each

D
Zero

Solution

(C) is correct
Question 3
1 / -0

To find mean from grouped data on the formula $$ \bar{x} = a + \frac{\sum f_{i}d_{i}}{ \sum f_{i}} d_{i}, $$ is deviation of following from a.

A
Lower limits of classes

B
Upper limits of classes

C
Mid-points of classes

D
Frequencies of group sign

Solution
Question 4
1 / -0

Number of students in a school according to their age are as follows :
Age (in yrs.) No. of students
8 15
9 25
10 40
11 36
12 41
13 37
14 20
15 13
16 5
17 3
Their mode is

A
41

B
12

C
3

D
17

Solution

From above table it is clear that frequency 41 is maximum and its mavimum and its corresponding age group of 12 . So its mode will be 12.
Thus (B) is correct.
Question 5
1 / -0

The mode of the data$$4, 3, 4, 5, 4, 2, 4, 1$$ will be :

A
$$1$$

B
$$2$$

C
$$5$$

D
$$4$$

Solution

Arranging the numbers with the same values together, we get
$$1,2,3,4,4,4,4,5$$
Mode of this data is $$4$$ because it occurs more frequently than other observations.
Question 6
1 / -0

The exam scores of all $$500$$ students were recorded and it was determined that these scores were normally distributed. If Jane's score is $$0.8$$ standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?(Area under the curve below $$z=0.8 \ is \ 0.7881$$)

A
$$394$$

B
$$250$$

C
$$400$$

D
$$106$$

Solution

Let the Jane's score be $$x$$.

Let $$m$$ be the mean and $$s$$ be the standard deviation and then we find the $$z$$ score.

Since it is given that Jane's score is $$0.8$$ standard deviation above the mean means $$x=0.8s+m$$

$$z=\dfrac { x-m }{ s } =\dfrac { 0.8s+m-m }{ s } =0.8\\$$
$$\Rightarrow z=0.8$$

The percentage of students who scored above Jane is (from table of normal distribution) is

$$1-0.7881=0.2119=21.19$$ $$\%$$

The number of students who scored above Jane is (from table of normal distribution) is

$$\dfrac { 21.9 }{ 100 } \times 500=106$$

Hence, $$106$$ students scored above Jane.
Question 7
1 / -0

Find the variance of the series 5, 8, 11, 14 and 17.....

A
17

B
18

C
16

D
12

Solution

Given series is $$ 5, 8, 11, 14, 17$$
Mean$$=\cfrac { 5+8+11+14+17 }{ 5 } =\cfrac { 55 }{ 5 } $$
Mean$$=\mu=11$$
Variance$$=\sum { { { \left( { x }_{ i }-\mu \right) }^{ 2 } }/{ n } } $$
$$=\cfrac { 1 }{ 5 } \left[ \left( 5-11 \right) ^{ 2 }+{ \left( 8-11 \right) }^{ 2 }+{ \left( 11-11 \right) }^{ 2 }+{ \left( 14-11 \right) }^{ 2 }+{ \left( 17-11 \right) }^{ 2 } \right] $$
$$=\cfrac { 1 }{ 5 } \left[ 36+9+0+9+36 \right] $$
$$=\cfrac { 90 }{ 5 } $$
$$=18$$
Question 8
1 / -0

The sum of the squares of deviations of 10 items about mean 50 is 250 .The coefficient of variation is

A
10%

B
50%

C
30%

D
none of these

Solution

$$\textbf{Step -1: Find the standard deviation.}$$
$$\text{It is given, }$$
$$\text{Mean, }\overline{x}=50,$$
$$n=10\text{ and }$$
$$\sum(x-\overline{x})^2=250$$
$$\therefore\text{Standard deviation}=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$$
$$=\sqrt{\dfrac{250}{10}}$$
$$=5$$
$$\textbf{Step -2: Find the coefficient of variance.}$$
$$\text{Coefficient of variation}=\dfrac{\text{Standard deviation}}{\text{Mean}}\times100\%.$$
$$=\dfrac{5}{50}\times100\%$$
$$=10\%$$
$$\textbf{Hence, option A is correct.}$$
Question 9
1 / -0

The probability distribution of a random variable $$X$$ is given below:
$$X=x$$ 0 1 2 3
$$P(X=x)$$ $$\frac{1}{10}$$ $$\frac{2}{10}$$ $$\frac{3}{10}$$ $$\frac{4}{10}$$
Then the variance of $$X$$ is

A
1

B
2

C
3

D
4

Solution

$$E[x^2]=0+1^2\cdot \displaystyle \frac{2}{10}+2^2\cdot \frac{3}{10}+3^2\cdot \frac{4}{10}=\frac{[2+12+36]}{10}=5.0$$
$$E[x]=0+\displaystyle \frac{2}{10}+2\cdot \frac{3}{10}+3\cdot \frac{4}{10}=\frac{[2+6+12]}{10}=2$$
$$\therefore$$ Var (x) $$= E[x^2] - E[x]^2 = 5-2^2=1$$
Question 10
1 / -0

The greatest value of a collection of data is $$72$$ and the least value is $$28$$. Then the coefficient of range is.

A
$$44$$

B
$$0.72$$

C
$$0.44$$

D
$$0.28$$

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Re-Attempt Weekly Quiz Competition

Age (in yrs.)	No. of students
8	15
9	25
10	40
11	36
12	41
13	37
14	20
15	13
16	5
17	3

$$X=x$$	0	1	2	3
$$P(X=x)$$	$$\frac{1}{10}$$	$$\frac{2}{10}$$	$$\frac{3}{10}$$	$$\frac{4}{10}$$

Statistics Test - 46

Result

Statistics Test - 46

Score

-

Rank

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SHARING IS CARING

Question 1

$$0.065$$

$$12.003$$

$$0.083$$

$$0.073$$

Solution

Question 2

Negative

Positve

Different in each

Zero

Solution

Question 3

Lower limits of classes

Upper limits of classes

Mid-points of classes

Frequencies of group sign

Solution

Question 4

41

12

3

17

Solution

Question 5

$$1$$

$$2$$

$$5$$

$$4$$

Solution

Question 6

$$394$$

$$250$$

$$400$$

$$106$$

Solution

Question 7

17

18

16

12

Solution

Question 8

10%

50%

30%

none of these

Solution

Question 9

1

2

3

4

Solution

Question 10

$$44$$

$$0.72$$

$$0.44$$

$$0.28$$

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