Statistics Test

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Statistics Test - 35

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Weekly Quiz Competition

Question 1

1 / -0

Calculate the standard deviation of the following data.
$$10, 20, 15, 8, 3, 4$$

$$5.97$$

$$59.7$$

$$4.97$$

None of these

Solution

$$\Rightarrow$$ The given numbers are $$3,4,8,10,15,20$$.

$$\Rightarrow$$ $$\overline{X}=\dfrac{3+4+8+10+15+20}{6}=10$$

$$X$$	$$X-\overline{X}$$	$$(X-\overline{X})^2$$
$$3$$	$$-7$$	$$49$$
$$4$$	$$-6$$	$$36$$
$$8$$	$$-2$$	$$4$$
$$10$$	$$0$$	$$0$$
$$15$$	$$5$$	$$25$$
$$20$$	$$10$$	$$100$$
		$$\sum (X-\overline{X})^2=214$$

$$\Rightarrow$$ $$S=\sqrt{\dfrac{(X-\overline{X})^2}{N}}$$

$$=\sqrt{\dfrac{214}{6}}$$

$$=\sqrt{35.66}$$

$$=5.97$$

Question 2
1 / -0

Which one of the following is a measure of dispersion?

A
Mean

B
Median

C
Mode

D
Standard deviation

Solution

Dispersion is the extent to which the distribution is stretched or squeezed.
The most common measures of dispersion are variance, standard deviation, mean deviation, range etc.
Question 3
1 / -0

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 4

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 5
1 / -0

If $$n=10, \bar{x}=12$$ and $$\sum x^2=1530$$, then calculate the coefficient of variation.

A
$$20$$

B
$$25$$

C
$$30$$

D
$$35$$

Solution

$$\sigma=\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$$

$$=\sqrt{\dfrac{1530}{10}-(12)^2}$$

$$=\sqrt{153-144}$$
$$=\sqrt{9}$$
$$=3$$

Coefficient of variation $$=\dfrac{\sigma}{\overline{x}}\times 100$$

$$=\dfrac{3}{12}\times 100$$

$$=\dfrac{1}{4}\times 100$$
$$=25$$

Question 6

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 7
1 / -0

The variance of $$20$$ observations is $$5$$. If each observation is multiplied by $$2$$, then what is the new variance of the resulting observations?

A
$$5$$

B
$$10$$

C
$$20$$

D
$$40$$

Solution

$$\textbf{Step -1: Write the formula of variance.}$$

$$\textbf{Variance = }$$$$\mathbf{ \dfrac{1}{n}\sum \left ( x_{i}-\bar{x} \right )^{2}}$$

$$\text{ where n is the number of observations}$$
$$x_{i}\text{ is the observations}$$
$$\bar{x}\text{ is mean value of the observations.}$$

$$\textbf{Step -2: Find }$$ $$\mathbf{\sum \left ( x_{i}-\bar{x} \right )^{2}}$$
$$\text{Let }x_{_1}, x_{_2}, ……x_{_{20}}\text{ are the observations.}$$
$$\text{We have, Variance = 5, n = 20}$$
$$\text{Variance} = \dfrac{1}{n}\sum \left ( x_{i}-\bar{x} \right )^{2}$$
$$\Rightarrow\dfrac{1}{20}\sum \left ( x_{i}-\bar{x} \right )^{2}=5$$
$$\Rightarrow\sum \left ( x_{i}-\bar{x} \right )^{2}= 100 \boldsymbol{\qquad\rightarrow(1)}$$

$$\textbf{Step -3: Write new observations.}$$
$$\text{Multiplying all observations with 2, we get new observations }y_{1}, y_{2}, ……y_{20}$$
$$\text{Where }y_{i}= 2x_{i} \boldsymbol{ \qquad\qquad→(2)}$$

$$\textbf{Step -4: Calculating the mean of y(new observations).}$$
$$\bar{y}=\dfrac{1}{n}\sum y_{i}$$
$$=\dfrac{1}{20}\sum 2x_{i}$$
$$=2\left \{ \left ( \dfrac{1}{20} \right )\sum x_{i} \right \}$$
$$=2\bar{x} \qquad\qquad \boldsymbol{ →(3})$$

$$\textbf{Step -5: Find }$$$$\mathbf{\sum \left ( y_{i}-\bar{y} \right )^{2}.}$$
$$\text{From equation 1, we can write,}$$$${\sum \left ( x_{i}-\bar{x} \right )^{2}=100}$$
$${\Rightarrow\sum \left ( \dfrac{1}{2}y_{i}-\dfrac{1}{2}\bar{y} \right )^{2}=100}$$ $$\textbf{[ From equation 2 and 3 ]}$$
$$\Rightarrow \left (\dfrac{1}{2} \right ) ^2\sum \left ( y_{i}-\bar{y} \right )^{2}=100$$
$$\Rightarrow \sum \left ( y_{i}-\bar{y} \right )^{2}=400$$

$$\textbf{Step - 6: Calculate new variance.}$$
$$\text{New variance}=\left (\dfrac{1}{n}\right )\sum \left ( y_{i}-\bar{y} \right )^{2}$$
$${=\dfrac{1}{20}\times 400=20}$$
$$\therefore\text{The new variance of resulting observation is 20.}$$
$$\textbf{Hence, the correct option is C.}$$
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

The variance of numbers $$x_1, x_2, x_3,..... x_n$$ is V. Consider the following statements :
1. If every $$x_1$$, is increased by 2, the variance of the new set of numbers is V.
2. If the numbers $$x_1$$ is squared, the variance of the new set is $$V^2$$.
Which of the following statements is/are correct ?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

In case 1, a deterministic value of 2 units is being added to every $$x_i$$ and hence, the variance of the set remains unchanged as $$V$$. New variance would be $$\dfrac{1}{n}\sum\left(x_i+2-\left(\sum\dfrac{x_i+2}{n}\right)\right)=\dfrac{1}{n}\sum\left(x_i-\sum\dfrac{x_i}{n}\right)=V$$
In case 2, each $$x_i$$ is squared and hence, the mean becomes $$\dfrac{1}{n}\sum x_i^2 $$. Hence, the new variance is $$\dfrac{1}{n}\sum\left( x_i^2-\dfrac{1}{n}\sum x_i^2\right)^2$$
and not $$V^2=\dfrac{1}{n^2}\sum\left( x_i-\dfrac{1}{n}\sum x_i\right)^4$$.
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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Re-Attempt Weekly Quiz Competition

Statistics Test - 35

Result

Statistics Test - 35

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SHARING IS CARING

Question 1

$$5.97$$

$$59.7$$

$$4.97$$

None of these

Solution

Question 2

Mean

Median

Mode

Standard deviation

Solution

Question 3

$$8$$ and $$20$$

$$8$$ and $$4$$

$$8$$ and $$8$$

$$80$$ and $$40$$

Solution

Question 4

$$416$$

$$46$$

$$16$$

None of these

Solution

Question 5

$$20$$

$$25$$

$$30$$

$$35$$

Solution

Question 6

$$6.32$$.

$$2.32$$.

$$3.32$$.

None of these

Solution

Question 7

$$5$$

$$10$$

$$20$$

$$40$$

Solution

Question 8

$$7.45$$

$$7.35$$

$$7.98$$

$$7.73$$

Solution

Question 9

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 10

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

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