Statistics Test

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Statistics Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The choices of the fruits of 42 students in a class are as follows:
A, O, B, M, A, G, B, G, A, G, B, M. A, O, M, A, B, G, M, B, A, O, M, O, G, B, O, M, G, A, A, B, M, O, M, G, B, A, M, O, M, O, where A, B, G, M and O stands for Apple, Banana, Grapes, Mango and Orange respectively.
Which two fruits are liked by an equal number of students?

A
Apple and Mango

B
Mango and Banana

C
Banana and Orange

D
Banana and Grapes

Solution

Number of students like Apple $$= 9$$
Number of students like Orange $$= 7$$
Number of students like Banana $$= 8$$
Number of students like Grapes $$= 8$$
Number of students like Mango $$= 10$$
So, Banana and Grapes are liked by equal number of students
Question 2
1 / -0

The average of five numbers is $$40$$ and the average of another six numbers is $$50$$. The average of all numbers taken together is _______.

A
$$44.44$$

B
$$45.00$$

C
$$45.45$$

D
$$90.00$$

Solution

The average of five numbers is 40 , so the total of these five numbers will be 5x40.
The average of another six numbers is 50 , so the total of these six numbers will be 6x50.
Therefore, the average of all these taken together will be
$$= \dfrac{( 200+ 300)}{5+6}$$
$$= \dfrac{500}{11 }$$
$$= 45.45 $$
Question 3
1 / -0

The average rainfall for a week, excluding Sunday, was $$10\ \text{cms}$$. Due to heavy rainfall on Sunday, the average for the week rose to $$15\ \text{cms}$$. How much rainfall was on Sunday?

A
$$25\ \text{cms}$$

B
$$35\ \text{cms}$$

C
$$45\ \text{cms}$$

D
$$55\ \text{cms}$$

Solution
Question 4
1 / -0

Variance of first $$20 $$ natural number is

A
$$\cfrac { 133 }{ 4 } $$

B
$$\cfrac { 379 }{ 12 } $$

C
$$\cfrac { 133 }{ 2 } $$

D
$$\cfrac { 399 }{ 4 } $$

Solution

Variance of first $$n$$ natural numbers is $$\displaystyle =\dfrac{\sum(x_i)^2}{n}-(Mean)^2$$
$$=\dfrac{1^2+2^2+3^2.....n^2}{n}-(\dfrac{n+1}{2})^2$$
$$\sigma^2=\dfrac{n(n+1)(2n+1)}{6n}-\dfrac{(n+1)^2}{4}$$
$$\boxed{\sigma^2=\dfrac{n^2-1}{12}}$$
So variance of first $$20$$ natural number
$$=\sigma^2=\dfrac{20^2-1}{12}=\dfrac{399}{12}=\dfrac{133}{4}$$
Question 5
1 / -0

The average marks of three batches of students having 70, 50 and 30 students respectively are 50, 55 and 45. Find the average marks of all the 150 students, taken together.

A
$$50.67$$

B
$$40.67$$

C
$$50.60$$

D
$$40.60$$

Solution

Given that, the average marks of 70, 50 and 30 students respectively are 50, 55 and 45 marks.
$$\therefore\ $$ the total marks of 70, 50 and 30 students will be,
$$70 \times 50, 50 \times 55$$ and $$30 \times 45$$ respectively.
The average marks of all the 150 students will be:
$$= \dfrac{ (3500 + 2750 + 1350 )}{( 70+50+30)}$$ $$\left[\text{Average}=\dfrac{\text{Sum of all quantities}}{\text{Number of quantities}}\right]$$
$$\Rightarrow \dfrac{7,600 }{150}$$
$$\Rightarrow 50.67$$

Hence, the average marks of all the 150 students, taken together are $$50.67$$
Question 6
1 / -0

If there are $$N$$ individual observations $$X1, X2, ....Xn$$, Arithmetic mean $$=$$?

A
$$\overline {X} = \dfrac {X_{1}\times X_{2} \times X_{3}\times ......X_{n}}{N}$$

B
$$\overline {X} = \dfrac {X_{1} + X_{2} + X_{3} +......X_{n}}{N}$$

C
$$\overline {X} \dfrac {\sum X_{i}}{N}$$

D
(B) or (C)

Solution

$$\overline {X} = \dfrac {X_{1} + X_{2} + X_{3} +......X_{n}}{N}=\dfrac{\sum{X}}{N}$$
This is the direct method of calculating arithmetic mean.
Question 7
1 / -0

If the two observations are $$10$$ and $$0$$, their arithmetic mean is _______.

A
$$10$$

B
$$0$$

C
$$5$$

D
none of the above

Solution

Arithmetic mean refers to the average amount in a given group of data. In this measure of central tendency, all the data are added up and then divided by the number of figures in the data in order to ascertain the mean.
Therefore,
Arithmetic mean $$= \dfrac{(10+0)}{2}$$

$$= \dfrac{10}{2}$$

$$= 5$$
Question 8
1 / -0

Number of Wicket taken by a team in series of $$10$$ matches are $$2, 3, 4, 5, 0, 1, 3, 3, 4, 3$$ find the mode of the set.

A
$$3$$

B
$$2.8$$

C
$$3.3$$

D
$$2.9$$

Solution

Mode is the highest occurring figure in a series. It is the value in a series of observation that repeats maximum number of times and which represents the whole series as most of the values in the series revolves around this value.
Since, 3 is occurring the highest number of times. Therefore, 3 is the mode.
Question 9
1 / -0

What is the mode value for the numbers $$5, 8, 6, 4, 10, 15, 18, 10$$?

A
$$18$$

B
$$10$$

C
$$14$$

D
None of above

Solution

Mode is the highest occurring figure in a series. It is the value in a series of observation that repeats maximum number of times and which represents the whole series as most of the values in the series revolves around this value. Since in the given series, 10 is occurring the highest number of times. Therefore, 10 is the mode of the series of given observations.
Question 10
1 / -0

Consider the following observation relating to marks obtained by $$10$$ students in CS Foundation Exam: $$201, 202, 207, 222, 276, 201, 246, 212, 285, 312$$.
Arithmetic mean $$=$$?

A
$$206$$

B
$$237.3$$

C
$$217.2$$

D
$$216.1$$

Solution

Mean refers to the average amount in a given group of data. In this measure of central tendency, all the data are added up and then divided by the number of figures in the data in order to ascertain the mean.

Mean $$\dfrac{(201+ 202+207+222+276+201+246+212+285+312}{10}$$
$$= 236.4 $$

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Re-Attempt Weekly Quiz Competition

Statistics Test - 39

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Statistics Test - 39

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SHARING IS CARING

Question 1

Apple and Mango

Mango and Banana

Banana and Orange

Banana and Grapes

Solution

Question 2

$$44.44$$

$$45.00$$

$$45.45$$

$$90.00$$

Solution

Question 3

$$25\ \text{cms}$$

$$35\ \text{cms}$$

$$45\ \text{cms}$$

$$55\ \text{cms}$$

Solution

Question 4

$$\cfrac { 133 }{ 4 } $$

$$\cfrac { 379 }{ 12 } $$

$$\cfrac { 133 }{ 2 } $$

$$\cfrac { 399 }{ 4 } $$

Solution

Question 5

$$50.67$$

$$40.67$$

$$50.60$$

$$40.60$$

Solution

Question 6

$$\overline {X} = \dfrac {X_{1}\times X_{2} \times X_{3}\times ......X_{n}}{N}$$

$$\overline {X} = \dfrac {X_{1} + X_{2} + X_{3} +......X_{n}}{N}$$

$$\overline {X} \dfrac {\sum X_{i}}{N}$$

(B) or (C)

Solution

Question 7

$$10$$

$$0$$

$$5$$

none of the above

Solution

Question 8

$$3$$

$$2.8$$

$$3.3$$

$$2.9$$

Solution

Question 9

$$18$$

$$10$$

$$14$$

None of above

Solution

Question 10

$$206$$

$$237.3$$

$$217.2$$

$$216.1$$

Solution

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