Statistics Test

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Statistics Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

Let $$x_i$$ represents the outcome on a fair die and $$f_i$$ be the corresponding frequency. The variance for random variable $$x_i$$ with following frequency distribution, is
$$x_i$$ 1 2 3 4 5 6
$$f_i$$ 1 2 3 4 5 6

A
$$2$$

B
$$3$$

C
$$\dfrac{28}{9}$$

D
$$\dfrac{20}{9}$$
Question 2
1 / -0

If the sum of mean and variance of binomial distribution is 4.8 for 5 trials then

A
mean is 3

B
variance is 0.6

C
mean is 3.2

D
variance is 0.8

Solution
Question 3
1 / -0

The mean deviation about the mean of the set of first $$n$$ natural numbers when $$n$$ is an odd number.

A
$$\dfrac{n^{2}-1}{4n}$$

B
$$\dfrac{n}{4}$$

C
$$\dfrac{n^{2}+1}{4n}$$

D
$$\dfrac{n^{2}-1}{12}$$

Solution

$$\overline { X } =\cfrac { \sum _{ i=1 }^{ n }{ i } }{ n } =\cfrac { n(n+1) }{ 2n } =\cfrac { n+1 }{ 2 } $$
$$\left| { d }_{ i } \right| =\left| r-\left( \cfrac { n+1 }{ 2 } \right) \right| $$$
$$\sum { \left| { d }_{ i } \right| } =\sum _{ r=1 }^{ n }{ \left| r-\left( \cfrac { n+1 }{ 2 } \right) \right| } =\sum _{ r=1 }^{ \cfrac { n+1 }{ 2 } }{ \cfrac { n+1 }{ 2 } } =r+\sum _{ r=\cfrac { n+1 }{ 2 } }^{ n }{ r-\left( \cfrac { n+1 }{ 2 } \right) } $$
$$\Rightarrow \sum { = } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) -\cfrac { 1 }{ 2 } \left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+3 }{ 2 } \right) -\left( \cfrac { n+1 }{ 2 } \right) \left( \cfrac { n+1 }{ 2 } \right) +\cfrac { n+1 }{ 4 } \left( \cfrac { n+1 }{ 2 } +n \right) \quad $$
$$\Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } \right) -\cfrac { \left( n+1 \right) \left( n+3 \right) }{ 8 } \Rightarrow \cfrac { n+1 }{ 4 } \left( \cfrac { 3n+1 }{ 2 } -\cfrac { n+3 }{ 2 } \right) $$
$$\cfrac { n+1 }{ 4 } \left( \cfrac { 2n-2 }{ 2 } \right) =\cfrac { { n }^{ 2 }-1 }{ 4 } $$
Question 4
1 / -0

Let $$\sigma^{2}$$ is variance of following frequency distribution
$$x_{1}$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$
$$f_{1}$$ $$1$$ $$0$$ $$1$$ $$7$$ $$9$$ $$4$$ $$1$$ $$1$$ $$1$$
then $$\sigma^{2}$$ is equal to

A
$$2.4$$

B
$$2.5$$

C
$$2.6$$

D
$$2.7$$
Question 5
1 / -0

The sum of first $$n$$ natural numbers is given by the expression $$(2n^{2}+3n)$$. The mean of the given numbers is

A
$$\left(\dfrac{2}{n}+3\right)$$

B
$$\left(2+\dfrac{3}{n}\right)$$

C
$$(2+3n)$$

D
$$(2n+3)$$

Solution
Question 6
1 / -0

If the sum and sum of squares of $$10$$ observations are $$12$$ and $$18$$ resp., then, The $$S.D$$ of observations is :-

A
$$\dfrac{1}{5}$$

B
$$\dfrac{2}{5}$$

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

D
$$\dfrac{4}{5}$$

Solution

$$\sum x=12,\sum x^2=18,N=10$$
$$SD=\sqrt{\cfrac{\sum x^2}{N}-(\cfrac{\sum x}{N})^2}$$
$$\implies SD=\sqrt{\cfrac{18}{10}-(\cfrac{12}{10})^2}=\cfrac{3}{5}$$
Question 7
1 / -0

The mean and variance of a random variable having a binomial distribution are $$4$$ and $$2$$ respectively , then $$P(X=1)$$ is

A
$$1/32$$

B
$$1/16$$

C
$$1/8$$

D
$$1/4$$

Solution
Question 8
1 / -0

The mean and the standard devition (s.d) of five observations are $$9$$ and $$0$$, respecively. If one of the observations is changed such that the mean of the new set of five obervatons becomes $$10$$, then their s. d. is:

A
0

B
1

C
2

D
4

Solution

Since the standard deviation before the change was $$0$$, all the observation was the mean, or $$9$$. Since one observation was changed and the new mean is $$10$$, we have the following equation.

$$ \dfrac{9+9+9+9+x}{5}=10 $$

$$ \Rightarrow \dfrac{36+x}{5}=10 $$

$$ \Rightarrow 36+x=50 $$

$$ \Rightarrow x=14 $$

The changed observation is$$14$$ .

All the observations are$$\left\{ 9,9,9,9,14 \right\}$$

Since the mean is $$10$$,

The variance is

$$ \dfrac{\left[ {{\left( 9-10 \right)}^{2}}+{{\left( 9-10 \right)}^{2}}+{{\left( 9-10 \right)}^{2}}+{{\left( 9-10 \right)}^{2}}{{\left( 14-10 \right)}^{2}} \right]}{5} $$

$$ \Rightarrow \dfrac{\left( 1+1+1+1+16 \right)}{5} $$

$$ \Rightarrow 4 $$

Hence, the standard deviation is $$\sqrt{4}=2$$
Question 9
1 / -0

Let $$x_{1},\ x_{2},...x_{100}$$ are $$100$$ above observation such that $$\displaystyle \sum { { x }_{ 1 }=0 } ,\ \displaystyle \sum _{ 1\le i<j\le 100 }^{ }{ \left| { x }_{ i }{ x }_{ j } \right| =80000 } $$ & mean deviation from their mean is $$5$$, then their standard deviation, is-

A
$$10$$

B
$$30$$

C
$$40$$

D
$$50$$
Question 10
1 / -0

Mean derivation of the numbers $$1, 2, 3,...$$

A
$$\dfrac{n(n + 1)}{2n + 1}$$

B
$$\dfrac{n(n - 1)}{2n + 1}$$

C
$$\dfrac{n(n^2 - 1)}{2n + 1}$$

D
None of these

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Re-Attempt Weekly Quiz Competition

$$x_{1}$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$	$$6$$	$$7$$	$$8$$	$$9$$
$$f_{1}$$	$$1$$	$$0$$	$$1$$	$$7$$	$$9$$	$$4$$	$$1$$	$$1$$	$$1$$

$$x_i$$	1	2	3	4	5	6
$$f_i$$	1	2	3	4	5	6

Statistics Test - 42

Result

Statistics Test - 42

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$\dfrac{28}{9}$$

$$\dfrac{20}{9}$$

Question 2

mean is 3

variance is 0.6

mean is 3.2

variance is 0.8

Solution

Question 3

$$\dfrac{n^{2}-1}{4n}$$

$$\dfrac{n}{4}$$

$$\dfrac{n^{2}+1}{4n}$$

$$\dfrac{n^{2}-1}{12}$$

Solution

Question 4

$$2.4$$

$$2.5$$

$$2.6$$

$$2.7$$

Question 5

$$\left(\dfrac{2}{n}+3\right)$$

$$\left(2+\dfrac{3}{n}\right)$$

$$(2+3n)$$

$$(2n+3)$$

Solution

Question 6

$$\dfrac{1}{5}$$

$$\dfrac{2}{5}$$

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

$$\dfrac{4}{5}$$

Solution

Question 7

$$1/32$$

$$1/16$$

$$1/8$$

$$1/4$$

Solution

Question 8

0

1

2

4

Solution

Question 9

$$10$$

$$30$$

$$40$$

$$50$$

Question 10

$$\dfrac{n(n + 1)}{2n + 1}$$

$$\dfrac{n(n - 1)}{2n + 1}$$

$$\dfrac{n(n^2 - 1)}{2n + 1}$$

None of these

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