Descriptive Statistics Test 27

Result

Descriptive Statistics Test 27

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Weekly Quiz Competition

Question 1
1 / -0

Find the mean for the following data using step deviation method.

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

Answer:- By shortcut Method
Class interval width (w) = 2

$$X_i$$ $$F_i$$ $$d=\cfrac{X-A}{i}$$ $$F_i d_i$$
2 10 -1 -10
4=A 20 0 0
6 30 1 30
8 40 2 80
$$\Sigma f = 100$$ $$\Sigma fd = 100$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times w=4+\cfrac{100}{100} \times 2 ={4+2}=6$$
C)6
Question 2
1 / -0

From the bar graph What is total no. of students in class VIII?

A
$$45$$

B
$$185$$

C
$$215$$

D
$$245$$

Solution

As seen in the graph, we can say the following things.
The total number of students in class VIII are $$40+48+52+45+30=215$$.
Hence, option C is correct.
Question 3
1 / -0

From the given bar graph In which year the difference of the values of export and import is maximum?

A
$$1986 - 87$$

B
$$1982 - 83$$

C
$$1984 - 85$$

D
$$1983 - 84$$

Solution

We can see the graph of years versus Export and Import.
The difference of the values of export and import is maximum in the year $$1986-87$$.
Question 4
1 / -0

From the given bar graph. In which year the export is minimum

A
$$1984 - 85$$

B
$$1986 - 87$$

C
$$1983 - 84$$

D
$$1982 - 83$$

Solution

The graph of years versus Export and Import in shown on $$x$$ and $$y$$ axis respectively.
As seen in the graph, we can say in year $$1982-83$$, the export is minimum.
Question 5
1 / -0

Choose the formula used for arithmetic mean of grouped data by step deviation method is

A
$$\bar { x } =A-\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

B
$$\bar { x } =A\times \dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

C
$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

D
$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } +i$$

Solution

The formula used for arithmetic mean of groped data by step deviation method is $$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$
$$A =$$ Assumed mean of the given data
$$\sum { f } =$$ Summation of the frequencies given in the grouped data
$$\sum { fd\prime } =$$ Summation of the frequencies and deviation of a given mean data
$$d\prime =\dfrac { x-A }{ i } $$
$$i =$$ Class interval width
$$\bar { x } =$$ Arithmetic mean
Question 6
1 / -0

From the bar graph, answer the following question What is the total no. of students in section $$A, B$$ and $$C$$`.

A
$$40$$

B
$$140$$

C
$$52$$

D
$$215$$

Solution

As seen from the graph, the total number of students in sections $$A,B$$ and $$C$$ are $$A+B+C$$
$$=40+48+52=140$$
Hence, option B is correct.
Question 7
1 / -0

Study the bar graph and answer the question given below:
What was the percentage decline in the production of oil from $$2007$$ to $$2008$$?

A
$$30\%$$

B
$$33\%$$

C
$$33.33\%$$

D
$$36.3\%$$

Solution

Percentage we can get easily by getting the difference between two and then by finding that difference is what percent of the initial value.
So Percentage $$= \dfrac{20 - 30}{30}\times 100\%$$
$$= -33.33%$$
So there was decline of $$33.33\%$$ from the year $$2007$$ and $$2008$$
Question 8
1 / -0

Study the bar graph and answer the question given below:
In how many of the given years was the production of oil more than average production of the given years?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Average production (in $$1000$$ tonnes) over the given years will be
$$\cfrac{1}{7}[15 + 30 + 20 + 40 + 50 + 30 + 60]$$
= $$\cfrac{245}{7} = 35$$
So, the production in the years $$2009, 2010, 2012$$ are more than the average production.
Question 9
1 / -0

Study the above graph and calculate the difference between the number of cars manufactured by company $$Y$$ in $$2001$$ and $$2000$$?

A
$$20000$$

B
$$30000$$

C
$$40000$$

D
$$50000$$

Solution

Number of cars manufactured by company $$Y$$ in $$2001$$ $$=80000$$
Number of cars manufactured by company $$Y$$ in $$2000$$ $$= 100000$$
The difference is $$ 100000-80000 = 20000$$
Question 10
1 / -0

Find the mean for the following data using step deviation method.

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Answer:- By shortcut Method
Class interval width (w) = 3

X F $$d=\cfrac{X-A}{i}$$ Fd
3 5 -1 -5
6=A 10 -0 -0
9 15 1 15
12 20 2 40
$$\Sigma f = 50$$ $$\Sigma fd = 50$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=6+\left(\cfrac{50}{50} \times 3\right) = 6+3 = 9$$
A) 9