Statistics Test

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Statistics Test - 31

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Weekly Quiz Competition

Question 1

1 / -0

Find the mean for the following data using step deviation method.

$$15.833$$

$$16.833$$

$$17.833$$

$$18.833$$

Solution

Answer:- By shortcut Method

Class interval width (i) = $$5-0 = 5$$

X	C. I. mid point	F	$$d=\cfrac{X-A}{i}$$	Fd
0-5	2.5	4	-1	-4
5-10	7.5=A	8	0	0
10-15	12.5	12	1	12
15-20	17.5	16	2	32
20-25	22.5	20	3	60
		$$\Sigma F = 60$$		$$\Sigma Fd=100$$

Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=7.5+\left(\cfrac{100}{60} \times 5\right)={7.5+8.33}=15.83$$

A) $$15.83$$

Question 2
1 / -0

A random variable $$X$$ has the probability distribution given below. Its variance is
X 1 2 3 4 5
P(X=x) k 2k 3k 2k k

A
$$\dfrac{16}{3}$$

B
$$\dfrac{4}{3}$$

C
$$\dfrac{5}{3}$$

D
$$\dfrac{10}{3}$$

Solution

We know that
$$\sum P(x) = 1$$
$$k+2k+3k+2k+k=1$$
$$9k=1$$
$$k=\frac{1}{9}$$
Now, var $$\sigma^2 = \sum xi^2P(xi)-(\sum\,xiP(xi))^2$$
$$=1(k)+2^2(2k)+3^2(3k)+4^2(2k)+5^2(k)-[1(k)+2(2k)+3(3k)+4(2k)+5(k)]^2$$
$$=93k-(27k)^2$$
$$=93(\cfrac{1}{9})-[27(\frac{1}{9})]^2$$
$$=\cfrac{31}{3}-3^2$$
$$=\cfrac{31-27}{3}$$
$$=\cfrac{4}{3}$$
Question 3
1 / -0

Brian got grades of $$92,89$$ and $$86$$ on his first three math tests. What grade must he get on his final test to have an overall average of $$90$$?

A
90

B
93

C
87

D
85

Solution

Let the grade he got on final test be $$x$$
The average of all four grades is $$90$$.
So, we have $$\dfrac {(92+89+86+x)}{4}=90$$
$$\Rightarrow x = 360-267 = 93$$
Question 4
1 / -0

The mean of a finite set X of numbers is $$14$$, the median of this set of numbers is $$12$$, and the standard deviation is $$1.8$$. A new set Y is formed by multiplying each member of the set S by $$3$$. Determine the correct statements w.r.t. set Y:
I. The mean of the numbers is set Y is $$42$$
II. The median of the numbers in set Y is $$36$$
III. The standard deviation of the numbers is set Y is $$5.4$$

A
I only

B
II only

C
I and II only

D
I and III only

E
I, II, and III

Solution

For set $$x$$
Mean $$=14$$
Median$$=12$$
S.D.$$=\sqrt { \sigma } $$
$$=1.8$$.

For set $$y$$
Mean $$=14\times 3= 42$$.
Median $$=12\times 3=36$$.
SD$$=1.8\times 3= 5.4$$.

Hence all three are correct.
Question 5
1 / -0

Grades for the test on proofs did not go as well as the teacher had hoped. The mean grade was 68, the median grade was 64, and the standard deviation was 12. The teacher curves the score by raising each score by a total of 7 points. Which of the following statements is true?
I. The new mean is 75.
II. The new median is 71.
III. The new standard deviation is 7.

A
I only

B
III only

C
I and II only

D
I, II, and III

E
None of the statements are true

Solution

If each observation is increased by $$7$$ then the mean will also be increased by $$7$$.
New Mean = Old Mean $$+7=68+7=75$$

If each observation is increased by $$7$$ then the median will also be increased by $$7$$.
New median = Old median$$+7=64+7=71$$

If each observation is increased/decreased by any quantity the standard deviation will remain same.
New standard deviation= old standard deviation $$=12$$
Question 6
1 / -0

The mean of $$48, 27, 36, 24, x$$, and $$2x$$ is $$37.x =$$

A
$$13\dfrac {1}{3}$$

B
$$16\dfrac {2}{3}$$

C
$$29$$

D
$$33\dfrac {3}{4}$$

E
$$40$$

Solution

Mean $$=$$ $$\dfrac{sum \quad of \quad no.}{total \quad no}$$
$$\Rightarrow 37=\dfrac{48+27+36+24+x+2x}{6}$$
$$\Rightarrow 135+3x=222$$
$$\Rightarrow 3x=222-135$$
$$\Rightarrow 3x=87$$
$$\Rightarrow x=\dfrac{87}{3}=29$$
Question 7
1 / -0

The average of six numbers is 4. If the average of two of those numbers is 2, what is the average of the other four numbers?

A
5

B
6

C
7

D
8

E
9

Solution
Question 8
1 / -0

Find the arithmetic mean of $$\dfrac {1}{2}, \dfrac {1}{3}, 2n$$ and $$m$$.

A
$$\dfrac {5 + 12n + 6m}{24}$$

B
$$\dfrac {5 + 8n + 4m}{24}$$

C
$$\dfrac {5 + 2n + m}{4}$$

D
$$\dfrac {5 + 12n + 6m}{6}$$

E
$$\dfrac {5 + 2n + m}{12}$$

Solution

The arithmetic mean of $$\dfrac {1}{2}, \dfrac {1}{3}, 2n, m$$ is
$$\dfrac { \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 3 } +2n+m }{ 4 } $$
$$=\dfrac { \dfrac { 5 }{ 6 } +2n+m }{ 4 } $$
$$=\dfrac { 5+12n+6m }{ 24 } $$
Question 9
1 / -0

Directions For Questions

Refer to the data given in the table below:
$$X:$$ $$3$$ $$5$$ $$12$$ $$3$$ $$2$$

...view full instructions

Find the coefficient of variation.

A
$$72.66$$

B
$$81.24$$

C
$$264$$

D
$$330$$

E
None

Solution

$$ \quad X\\ \quad 03\\ \quad 05\\ \quad 12\\ \quad 03\\ \quad 02\\ \_ \_ \_ \_ \_ \\ \quad 25$$ $$x-\overline { x } \\ -2\\ \quad 00\\ \quad 07\\ -2\\ -3 $$ $${ \quad \quad (x-\overline { x } ) }^{ 2 }\\ \quad \quad \quad 04\\ \quad \quad \quad 00\\ \quad \quad \quad 49\\ \quad \quad \quad 04\\ \quad \quad \quad 09\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \sum { { (x-\overline { x } ) }^{ 2 }=66 } $$

Variance $$= \cfrac { \sum { { (x-\overline { x } ) }^{ 2 } } }{ n-1 } \\ =\cfrac { 66 }{ 4 } \\ =16.5$$.

$$\sigma =S.D.=\sqrt { 16.5 } $$.
Coefficient of variation $$= \cfrac { \sigma }{ x } \\ =\cfrac { \sqrt { 16.5 } }{ 5 } \\ =0.8124$$.

% of coefficient of variation $$= 0.8124*100\\ =81.24$$.
Question 10
1 / -0

Which of the following can be used as measures of dispersions?

A
Range

B
Percentiles

C
Standard Deviation

D
All

Solution

Following terms used as measures of dispersion are:-

a) Range
b) Percentile
c) Standard eviation
d) Mean deviation.

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X	1	2	3	4	5
P(X=x)	k	2k	3k	2k	k

Statistics Test - 31

Result

Statistics Test - 31

Score

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Rank

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SHARING IS CARING

Question 1

$$15.833$$

$$16.833$$

$$17.833$$

$$18.833$$

Solution

Question 2

$$\dfrac{16}{3}$$

$$\dfrac{4}{3}$$

$$\dfrac{5}{3}$$

$$\dfrac{10}{3}$$

Solution

Question 3

90

93

87

85

Solution

Question 4

I only

II only

I and II only

I and III only

I, II, and III

Solution

Question 5

I only

III only

I and II only

I, II, and III

None of the statements are true

Solution

Question 6

$$13\dfrac {1}{3}$$

$$16\dfrac {2}{3}$$

$$29$$

$$33\dfrac {3}{4}$$

$$40$$

Solution

Question 7

5

6

7

8

9

Solution

Question 8

$$\dfrac {5 + 12n + 6m}{24}$$

$$\dfrac {5 + 8n + 4m}{24}$$

$$\dfrac {5 + 2n + m}{4}$$

$$\dfrac {5 + 12n + 6m}{6}$$

$$\dfrac {5 + 2n + m}{12}$$

Solution

Question 9

$$72.66$$

$$81.24$$

$$264$$

$$330$$

None

Solution

Question 10

Range

Percentiles

Standard Deviation

All

Solution

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