Descriptive Statistics Test 32

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Descriptive Statistics Test 32

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Weekly Quiz Competition

Question 1
1 / -0

Variance of the first $$11$$ natural numbers is:

A
$$\sqrt{5}$$

B
$$\sqrt{10}$$

C
$$5\sqrt{2}$$

D
$$10$$

Solution

Sum of first $$n$$ natural numbers is $$\dfrac {n^2-1}{12}$$
$$=\dfrac {11^2-1}{12}$$
$$=\dfrac {120}{12}$$
$$=10$$
Thus, variance is $$\sqrt {\dfrac {n^2-1}{12}}$$ $$=\sqrt {10}$$.
Question 2
1 / -0

If $$x $$ is multiplied by $$k$$ then $$\sigma$$ changes to

A
$$k^2\sigma$$

B
$$k\sigma$$

C
$$k\sqrt{\sigma}$$

D
$$k(\sigma)^2$$

Solution

By the properties of standard deviation,
$$\sigma(kx)=|k|\sigma(x)$$
Therefore, by multiplying the $$x$$ by $$k$$, SD changes to $$k\sigma$$
Question 3
1 / -0

Standard Deviation of n observations $$a_1, a_2, a_3 .....a_n$$ is $$\sigma$$ Then the standard deviation of the observations $$\lambda a_1, \lambda a_2, ..., \lambda a_n$$ is

A
$$\lambda \sigma$$

B
$$-\lambda \sigma$$

C
$$|\lambda| \sigma$$

D
$${\lambda}^n \sigma$$

Solution

Let the mean of original data is $$\mu$$
Then mean of observation if all the terms are multiplied by the constant $$\lambda$$ will be $$\lambda \mu$$
If $$\sigma$$ be the standard deviation of original observation , then
$$\sigma=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n (x_i-\mu)^2 }$$
Now let the standard deviation of new observation is $$\sigma'$$, then
$$\sigma'=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n (\lambda x_i-\lambda\mu)^2 }$$
$$=\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n \lambda^2( x_i-\mu)^2 }$$
$$=|\lambda|\sqrt{\dfrac{1}{n}\displaystyle \sum_{i=1}^n( x_i-\mu)^2 }=|\lambda |\sigma$$
Note that: standard deviation can never be negative
Question 4
1 / -0

The standard deviation of $$a,a+d,a+2d,...,a+{2}{n}d$$ is

A
$$nd$$

B
$${n}^{2}d$$

C
$$\sqrt { \cfrac { n(n+1) }{ 3 } } d\quad $$

D
$$\sqrt { \cfrac { n(n+3) }{ 3 } } d\quad $$

Solution

Series is $$a, a+d, a+2d,\cdots, a+2nd$$
Total number of terms: $$N = 2n+1$$
Sum of the series: $$S = \cfrac{2n+1}2(a+a+2nd) = (2n+1)(a+nd)$$
Mean of the data: $$\bar{x} = \cfrac SN = \cfrac{(2n+1)(a+nd)}{(2n+1)} = a+nd$$
Standard Deviation: $$s.d. = \sqrt{\cfrac{\displaystyle\sum_{i=1}^{2n+1} (x_i - \bar x)^2}{N}}$$
$$s.d. = \sqrt{\cfrac{2(1^2+2^2+\cdots+n^2)}{(2n+1)} = \cfrac{2\times\dfrac{n(n+1)(2n+1)}{6}}{(2n+1)}d^2}$$
$$s.d. = \sqrt{\cfrac{n(n+1)}{3}}d$$
Question 5
1 / -0

If the variance of a data is $$12.25$$, then the standard deviation is

A
$$3.5$$

B
$$3$$

C
$$2.5$$

D
$$3.25$$

Solution

$$S.D=\sqrt{variance}$$
$$\Rightarrow S.D=\sqrt{12.25}$$
$$\Rightarrow S.D=3.5$$
Option A is correct.

Question 6

1 / -0

Find the variance of the following distribution.

Class interval	$$20-24$$	$$25-29$$	$$30-34$$	$$35-39$$	$$40-44$$	$$45-49$$
Frequency	$$15$$	$$25$$	$$28$$	$$12$$	$$12$$	$$8$$

$$416$$

$$46$$

$$16$$

None of these

Solution

$$Class$$	$$Mid-point$$ $$(x)$$	$$Frequency$$ $$(f)$$	$$xf$$	$$(x-\overline{x})$$	$$(x-\overline{x})^2$$	$$(x-\overline{x})^2f$$
$$20-24$$	$$22$$	$$15$$	$$330$$	$$-10.25$$	$$105.0625$$	$$1575.9375$$
$$25-29$$	$$27$$	$$25$$	$$675$$	$$-5.25$$	$$27.5625$$	$$689.0625$$
$$30-34$$	$$32$$	$$28$$	$$896$$	$$-0.25$$	$$0.0625$$	$$1.75$$
$$35-39$$	$$37$$	$$12$$	$$444$$	$$4.75$$	$$22.5625$$	$$270.75$$
$$40-44$$	$$42$$	$$12$$	$$504$$	$$9.75$$	$$95.0625$$	$$1140.75$$
$$45-49$$	$$47$$	$$8$$	$$376$$	$$14.75$$	$$217.5625$$	$$1740.5$$
		$$\sum f=100$$	$$\sum xf=3225$$			$$\sum(x-\overline{x})^2f=5418.75$$

$$\Rightarrow$$ $$Mean(\overline{x})=\dfrac{\sum xf}{\sum f}=\dfrac{3225}{100}=32.25$$

$$\Rightarrow$$ $$Variance(\sigma^2)=\dfrac{\sum(x-\overline{x})^2f}{\sum f}=\dfrac{5418.75}{100}=54.18$$

Question 7
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The mean and the variance of $$10$$ observations are given to be $$4$$ and $$2$$ respectively. If every observation is multiplied by $$2$$, the mean and the variance of the new series will be respectively.

A
$$8$$ and $$20$$

B
$$8$$ and $$4$$

C
$$8$$ and $$8$$

D
$$80$$ and $$40$$

Solution

Mean is summation of all observations divided by number of observation and hence, would be multiplied by $$2$$ due to linearity. Hence, the new mean is $$8$$.
Variance is sum of squares of each observation subtracted by the mean. Hence, due to squared dependence, it will be quadrupled to get $$8$$.
Hence, both mean and variance are $$8$$.
Question 8
1 / -0

Directions For Questions

The mean and standard deviation of $$100$$ items are $$50, 5$$ and that of $$150$$ items are $$40, 6$$ respectively.

...view full instructions

What is the combined standard deviation of all 250 items ?

A
$$7.45$$

B
$$7.35$$

C
$$7.98$$

D
$$7.73$$

Solution

Here $$n_1=100,\bar{x}_1=50, s_1=5,n_2=150,\bar{x}_2=40,s_2=6$$

$$\therefore \sigma=\sqrt{\dfrac{n_1s_1^2+n_2s_2^2}{n_1+n_2}+\dfrac{n_1n_2(\bar{x}_1-\bar{x}_2)^2}{(n_1+n_2)^2}}$$

$$=\sqrt{\dfrac{100(5)^2+150(6)^2}{100+150}+\dfrac{(100)(150)(50-40)^2}{(100+150)^2}}$$

$$=\sqrt{\dfrac{158}{5}+24}$$

$$=\sqrt{31.6+24}$$

$$=\sqrt{55.6}$$

$$=7.45$$

$$\therefore \sigma=7.45$$

Question 9

1 / -0

Calculate the standard deviation of the following data.

x	$$3$$	$$8$$	$$13$$	$$18$$	$$23$$
f	$$7$$	$$10$$	$$15$$	$$10$$	$$8$$

$$6.32$$.

$$2.32$$.

$$3.32$$.

None of these

Solution

$$x$$	$$f$$	$$xf$$	$$(x-\overline{x})$$	$$(x-\overline{x})^2$$	$$(x-\overline{x})^2f$$
$$3$$	$$7$$	$$21$$	$$-10.2$$	$$104.04$$	$$728.28$$
$$8$$	$$10$$	$$80$$	$$-5.2$$	$$27.04$$	$$270.4$$
$$13$$	$$15$$	$$195$$	$$-0.2$$	$$0.04$$	$$0.6$$
$$18$$	$$10$$	$$180$$	$$4.8$$	$$23.04$$	$$230.4$$
$$23$$	$$8$$	$$184$$	$$9.8$$	$$96.04$$	$$768.32$$
	$$\sum f=50$$	$$\sum xf=660$$			$$\sum(x-\overline{x})^2=1998$$

$$\Rightarrow$$ $$\overline{x}=\dfrac{\sum xf}{\sum f}=\dfrac{660}{50}=13.2$$

Now, calculate standard deviation :

$$S=\sqrt{\dfrac{(x-\overline{x})^2}{\sum f}}=\sqrt{\dfrac{1998}{50}}=\sqrt{39.96}=6.32$$

Question 10
1 / -0

Let X denote the number of scores which exceed 4 in 1 toss of a symmetrical die. Consider the following statements :
1. The arithmetic mean of X is 6.
2. The standard deviation of X is 2.
Which of the above statements is/are correct ?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

The only possible favorable outcomes are $$5$$ and $$6$$. Hence, the arithmetic mean of $$X$$ is 5.5.
The standard deviation of $$X=\dfrac{1}{2}((6-5.5)^2+(5-55)^2)=0.25$$

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Descriptive Statistics Test 32

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Descriptive Statistics Test 32

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SHARING IS CARING

Question 1

$$\sqrt{5}$$

$$\sqrt{10}$$

$$5\sqrt{2}$$

$$10$$

Solution

Question 2

$$k^2\sigma$$

$$k\sigma$$

$$k\sqrt{\sigma}$$

$$k(\sigma)^2$$

Solution

Question 3

$$\lambda \sigma$$

$$-\lambda \sigma$$

$$|\lambda| \sigma$$

$${\lambda}^n \sigma$$

Solution

Question 4

$$nd$$

$${n}^{2}d$$

$$\sqrt { \cfrac { n(n+1) }{ 3 } } d\quad $$

$$\sqrt { \cfrac { n(n+3) }{ 3 } } d\quad $$

Solution

Question 5

$$3.5$$

$$3$$

$$2.5$$

$$3.25$$

Solution

Question 6

$$416$$

$$46$$

$$16$$

None of these

Solution

Question 7

$$8$$ and $$20$$

$$8$$ and $$4$$

$$8$$ and $$8$$

$$80$$ and $$40$$

Solution

Question 8

$$7.45$$

$$7.35$$

$$7.98$$

$$7.73$$

Solution

Question 9

$$6.32$$.

$$2.32$$.

$$3.32$$.

None of these

Solution

Question 10

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

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