Descriptive Statistics Test 46

Result

Descriptive Statistics Test 46

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Weekly Quiz Competition

Question 1
1 / -0

Observe the adjoining bar graph, showing the number of one-day international matches played by cricket teams of different countries. Choose the correct answer from the given four options:
How many more matches were played by India than Pakistan?

A
$$6$$

B
$$12$$

C
$$24$$

D
$$30$$

Solution

The given bar graph represents the number of matches played by cricket teams of different country.
We can prepare frequency table from given bar graph as follows:
Country No. of matches
India $$30$$
Pakistan $$24$$
West Indies $$20$$
England $$28$$
South Africa $$18$$
Australia $$32$$
Sri Lanka $$24$$
Total number of students $$=176$$.
Then, the number of more matches played by India than Pakistan
$$=$$ Number of matches played by India $$-$$ Number of matches played by Pakistan
$$=30-24=6$$ matches.
That is, India played $$6$$ matches more than Pakistan.
Hence, option $$A$$ is correct.
Question 2
1 / -0

Observe the following pictograph which shows the number of ice cream cones sold by school canteen during a week. Chose the correct answer from the given options:
If the cost of one ice cream cone is $$Rs. 20,$$ then the sale value on friday was:

A
Rs.$$70$$

B
Rs.$$140$$

C
Rs.$$280$$

D
Rs.$$1340$$

Solution

Number of ice creams sold on friday $$= 7 \times 2$$
$$ = 14 $$
$$\therefore$$ Sale value on Thursday $$= Rs.20 \times 14 $$
$$= Rs. 280$$
Question 3
1 / -0

Observe the adjoining bar graph, showing the number of one-day international matches played by cricket teams of different countries. Choose the correct answer from the given four options:
Which country played maximum number of matches?

A
India

B
England

C
Pakistan

D
Australia

Solution

The given bar graph represents the number of matches played by cricket teams of different country.
We can prepare frequency table from given bar graph as follows:
Country No. of matches
India $$30$$
Pakistan $$24$$
West Indies $$20$$
England $$28$$
South Africa $$18$$
Australia $$32$$
Sri Lanka $$24$$
Total number of students $$=176$$.
The country that played maximum number of matches will have the highest number of matches.
Clearly, the maximum number of matches are played by Australia $$(32)$$.
Hence, option $$D$$ is correct.
Question 4
1 / -0

Observe the adjoining bar graph, showing the number of one-day international matches played by cricket teams of different countries. Choose the correct answer from the given four options:
Ratio of the number of matches played by India to the number of matches played by Sri Lanka is:

A
$$4 : 5$$

B
$$5 : 4$$

C
$$4 : 3$$

D
$$7 : 6$$

Solution

The given bar graph represents the number of matches played by cricket teams of different country.
We can prepare frequency table from given bar graph as follows:
Country No. of matches
India $$30$$
Pakistan $$24$$
West Indies $$20$$
England $$28$$
South Africa $$18$$
Australia $$32$$
Sri Lanka $$24$$
Total number of students $$=176$$.
Then, the ratio of number of matches played by India to the number of matches played by Sri Lanka
$$=\dfrac{\text{Number of matches played by India}}{\text{Number of matches played by Sri Lanka}}$$
$$=\dfrac{30}{24}=\dfrac{5}{4}=5:4$$.
Hence, the required ratio is $$5:4$$.
Therefore, option $$B$$ is correct.
Question 5
1 / -0

In a series $$\sum x^{2} = 100, n = 5 $$ and $$\sum x = 20 $$ , then standard deviation is

A
16

B
2

C
4

D
8

Solution

Standard deviation
$$\sigma \sqrt{\dfrac{\sum x^{2}}{n} - \left ( \dfrac{\sum }{n} \right )^{2}} = \sqrt{\dfrac{100}{5} - \left ( \dfrac{20}{5} \right )^{2}} $$
$$\sqrt{20-16} = \sqrt{4} = 2 $$
Question 6
1 / -0

Standard deviation of data $$6, 10, 4, 7, 4, 5$$ is -

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

B
$$\dfrac{13}{3} $$

C
$$\sqrt{26} $$

D
$$\dfrac{\sqrt{26}}{6}$$

Solution

Mean $$(\bar{x}) = \dfrac{6+ 10 + 4 + 7 + 4 + 5}{6} = \dfrac{36}{6} = 6 $$
Standard deviation $$(\sigma ) = \sqrt{\dfrac{1}{N} \times \sum (x_{1} - \bar{x})^{2}} $$
$$=\sqrt{\dfrac{1}{6} \times 26} = \sqrt{\dfrac{13}{3}}$$
Question 7
1 / -0

The exam scores of all $$500$$ students were recorded and it was determined that these scores were normally distributed. If Jane's score is $$0.8$$ standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?(Area under the curve below $$z=0.8 \ is \ 0.7881$$)

A
$$394$$

B
$$250$$

C
$$400$$

D
$$106$$

Solution

Let the Jane's score be $$x$$.

Let $$m$$ be the mean and $$s$$ be the standard deviation and then we find the $$z$$ score.

Since it is given that Jane's score is $$0.8$$ standard deviation above the mean means $$x=0.8s+m$$

$$z=\dfrac { x-m }{ s } =\dfrac { 0.8s+m-m }{ s } =0.8\\$$
$$\Rightarrow z=0.8$$

The percentage of students who scored above Jane is (from table of normal distribution) is

$$1-0.7881=0.2119=21.19$$ $$\%$$

The number of students who scored above Jane is (from table of normal distribution) is

$$\dfrac { 21.9 }{ 100 } \times 500=106$$

Hence, $$106$$ students scored above Jane.
Question 8
1 / -0

Find the variance of the series 5, 8, 11, 14 and 17.....

A
17

B
18

C
16

D
12

Solution

Given series is $$ 5, 8, 11, 14, 17$$
Mean$$=\cfrac { 5+8+11+14+17 }{ 5 } =\cfrac { 55 }{ 5 } $$
Mean$$=\mu=11$$
Variance$$=\sum { { { \left( { x }_{ i }-\mu \right) }^{ 2 } }/{ n } } $$
$$=\cfrac { 1 }{ 5 } \left[ \left( 5-11 \right) ^{ 2 }+{ \left( 8-11 \right) }^{ 2 }+{ \left( 11-11 \right) }^{ 2 }+{ \left( 14-11 \right) }^{ 2 }+{ \left( 17-11 \right) }^{ 2 } \right] $$
$$=\cfrac { 1 }{ 5 } \left[ 36+9+0+9+36 \right] $$
$$=\cfrac { 90 }{ 5 } $$
$$=18$$
Question 9
1 / -0

Which two years did the least number of boys attend the convention?

A
1995 and 1996

B
1995 and 1998

C
1996 and 1997

D
1996 and 1992

E
1997 and 1998
Question 10
1 / -0

The probability distribution of a random variable $$X$$ is given below:
$$X=x$$ 0 1 2 3
$$P(X=x)$$ $$\frac{1}{10}$$ $$\frac{2}{10}$$ $$\frac{3}{10}$$ $$\frac{4}{10}$$
Then the variance of $$X$$ is

A
1

B
2

C
3

D
4

Solution

$$E[x^2]=0+1^2\cdot \displaystyle \frac{2}{10}+2^2\cdot \frac{3}{10}+3^2\cdot \frac{4}{10}=\frac{[2+12+36]}{10}=5.0$$
$$E[x]=0+\displaystyle \frac{2}{10}+2\cdot \frac{3}{10}+3\cdot \frac{4}{10}=\frac{[2+6+12]}{10}=2$$
$$\therefore$$ Var (x) $$= E[x^2] - E[x]^2 = 5-2^2=1$$

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$$X=x$$	0	1	2	3
$$P(X=x)$$	$$\frac{1}{10}$$	$$\frac{2}{10}$$	$$\frac{3}{10}$$	$$\frac{4}{10}$$

Descriptive Statistics Test 46

Result

Descriptive Statistics Test 46

Score

-

Rank

-

SHARING IS CARING

Question 1

$$6$$

$$12$$

$$24$$

$$30$$

Solution

Question 2

Rs.$$70$$

Rs.$$140$$

Rs.$$280$$

Rs.$$1340$$

Solution

Question 3

India

England

Pakistan

Australia

Solution

Question 4

$$4 : 5$$

$$5 : 4$$

$$4 : 3$$

$$7 : 6$$

Solution

Question 5

16

2

4

8

Solution

Question 6

$$\sqrt{\dfrac{13}{3}} $$

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

$$\sqrt{26} $$

$$\dfrac{\sqrt{26}}{6}$$

Solution

Question 7

$$394$$

$$250$$

$$400$$

$$106$$

Solution

Question 8

17

18

16

12

Solution

Question 9

1995 and 1996

1995 and 1998

1996 and 1997

1996 and 1992

1997 and 1998

Question 10

1

2

3

4

Solution

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