Descriptive Statistics Test 43

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Descriptive Statistics Test 43

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Weekly Quiz Competition

Question 1
1 / -0

Choose the correct answers from the alternatives given.
Directions for question 74 and 75: Study the following pie chart and answer the questions based on it.
The following pie-chart shows the result of an examination of 360 students.
The number of students who failed is

A
30

B
26

C
10

D
36

Solution

Number of the students failed = $$\frac{10}{100} \, \times$$ 360 = 36
Question 2
1 / -0

Let mean of $$100$$ data points be $$55$$ and variance be $$16$$ . Now if every data point is increased by $$2$$ units, then the sum of mean and standard deviation of new data point will be:

A
$$71$$

B
$$59$$

C
$$63$$

D
$$61$$

Solution

As the 2 units are added to every point on the data , there will be no change in the standard deviation .But mean will raise by 2units
Hence new mean = 57
Given Variance=16
Standard deviation = $$\sqrt{variance}=\sqrt{16}=4$$
Sum of mean and standard deviation = $$57+4=61$$ units
Question 3
1 / -0

How many teachers are more than $$35$$ years old?

A
$$3$$

B
$$5$$

C
$$10$$

D
$$2$$

Solution

$$\Rightarrow$$ From given histogram we are going to form frequency distribution table:

$$Ages$$
$$(Years)$$ $$No.\,of\,teachers$$
$$20-25$$ $$4$$
$$25-30$$ $$5$$
$$30-35$$ $$6$$
$$35-40$$ $$3$$
$$40-45$$ $$2$$
$$45-50$$ $$5$$
$$\Rightarrow$$ From above table we can see, the number of teachers more than $$35$$ years old $$=3+2+5=10$$
Question 4
1 / -0

For $$X \rightarrow B (n, p), \,$$ if $$\, n = 25, E (x) = 10$$ , then S.D. $$(x) = $$

A
$$2\sqrt{6}$$

B
$$2.4$$

C
$$2 \sqrt{10}$$

D
$$\sqrt{2.4}$$

Solution

In the given Binomial Distribution,
Number of trials, $$n=25$$
Let probability of success $$=p$$

Mean $$=E(x)=10$$

$$\Rightarrow np=10$$

$$25p=10$$

$$p=\cfrac{10}{25}=\cfrac{2}{5}$$

Then probability of failure, $$q=1-p=\cfrac{3}{5}$$

Standard deviation for Binomial Distribution $$=\sqrt{npq}$$
$$=\sqrt{25\times \cfrac{2}{5}\times \cfrac{3}{5}}$$
$$=\sqrt{6}$$
$$\approx 2.45$$

Question 5

1 / -0

Find variance of the following data.

Class interval	Frequency
$$4-8$$	$$3$$
$$8-12$$	$$6$$
$$12-16$$	$$4$$
$$16-20$$	$$7$$

$$19$$

$$20$$

$$21$$

$$23$$

Solution

$$C.I$$	$$x_i$$	$$f_i$$	$$f_ix_i$$	$$(x_i-\overline{x})^2$$	$$f_i(x_i-\overline{x})^2$$
$$4-8$$	$$6$$	$$3$$	$$18$$	$$49$$	$$147$$
$$8-12$$	$$10$$	$$6$$	$$60$$	$$9$$	$$54$$
$$12-16$$	$$14$$	$$4$$	$$56$$	$$1$$	$$4$$
$$16-20$$	$$18$$	$$7$$	$$126$$	$$25$$	$$175$$
		$$\sum f_i=20$$	$$\sum f_ix_i=260$$	$$\sum(x_i-\overline{x})^2=84$$	$$\sum f_i(x_i-\overline{x})^2=380$$

$$\Rightarrow$$ $$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{260}{20}=13$$

$$\Rightarrow$$ $$Variance(\sigma^2)\dfrac{\sum f_i(x_i-\overline{x})^2}{\sum f_i}=\dfrac{380}{20}=19$$

Question 6
1 / -0

If $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right) = 9}$$ and $$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$$, then the standard deviation of the $$9$$ items $${x_1},{x_2},.....,{x_9}$$ is

A
$$2$$

B
$$3$$

C
$$9$$

D
$$4$$

Solution

S.D of $$xi-5$$ is

$$\sigma =\sqrt{\dfrac{\sum_{i=1}^{9}(xi-5)^2}{9}-\left [ \dfrac{\sum_{i=1}^{9}(xi-5)^2}{9} \right ]^2}$$

$$\sigma =\sqrt{5-1}=2$$
Question 7
1 / -0

Suppose a population $$A$$ has $$100$$ observations $$101,102,........,200$$ and another population $$B$$ has $$100$$ observations $$151,152......,250 $$. If $$V_{A}and V_{B}$$ represent the variences of the two populations respectively, then $$V_{A}/ V_{B}$$ is

A
$$1$$

B
$$9/4$$

C
$$4/9$$

D
$$2/3$$

Solution
Question 8
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 9
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 10
1 / -0

Find the standard deviation of 10 observation 111,211,311,....1011.

A
100$$\sqrt { 3 } $$

B
250

C
300

D
50$$\sqrt { 3 } $$

Solution

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Re-Attempt Weekly Quiz Competition

$$Ages$$ $$(Years)$$	$$No.\,of\,teachers$$
$$20-25$$	$$4$$
$$25-30$$	$$5$$
$$30-35$$	$$6$$
$$35-40$$	$$3$$
$$40-45$$	$$2$$
$$45-50$$	$$5$$

Descriptive Statistics Test 43

Result

Descriptive Statistics Test 43

Score

-

Rank

-

SHARING IS CARING

Question 1

30

26

10

36

Solution

Question 2

$$71$$

$$59$$

$$63$$

$$61$$

Solution

Question 3

$$3$$

$$5$$

$$10$$

$$2$$

Solution

Question 4

$$2\sqrt{6}$$

$$2.4$$

$$2 \sqrt{10}$$

$$\sqrt{2.4}$$

Solution

Question 5

$$19$$

$$20$$

$$21$$

$$23$$

Solution

Question 6

$$2$$

$$3$$

$$9$$

$$4$$

Solution

Question 7

$$1$$

$$9/4$$

$$4/9$$

$$2/3$$

Solution

Question 8

mean is 3

variance is 0.6

mean is 3.2

variance is 0.8

Solution

Question 9

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

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

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

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

Solution

Question 10

100$$\sqrt { 3 } $$

250

300

50$$\sqrt { 3 } $$

Solution

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