Descriptive Statistics Test 28

Result

Descriptive Statistics Test 28

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$43$$

B
$$44$$

C
$$45$$

D
$$46$$

Solution

Answer:- By shortcut Method
Class interval width (i)= 24-12 = 12

X F $$d=\cfrac{x-A}{i} $$ Fd
12=A 1 0 0
24 2 1 2
36 3 2 6
48 4 3 12
60 5 4 20
$$\Sigma f = 15$$ $$\Sigma fd = 40$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=12+\left(\cfrac{40}{15} \times 12\right)={12+32}=44$$
B)44
Question 2
1 / -0

Study the above graph and calculate the difference between the number of cars manufactured by Company $$X$$ in $$2001$$ and $$2002$$?

A
$$20000$$

B
$$30000$$

C
$$40000$$

D
$$50000$$

Solution

Number of cars manufacturd by $$X$$ in $$2001 = 120000$$
Number of cars manufactured by $$ X$$ in $$2002=80000$$
The difference is $$120000-80000=40000$$
Question 3
1 / -0

Study the bar graph and answer the question given below:
What was the percent increase in production of oil in $$2006$$ compared to that in $$2012$$?

A
$$100\%$$

B
$$200\%$$

C
$$300\%$$

D
$$400\%$$

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 $$= \dfrac{60 - 15}{15}\times 100\%$$
$$= 300%$$
So there was an increase of $$300\%$$ from the year $$2006$$ and $$2012$$
Question 4
1 / -0

Study the bar graph and answer the question given below:
The average production of $$2007$$ and $$2009$$ was exactly to the average production of which of the following pair of the years?

A
$$2007$$ and $$2006$$

B
$$2007$$ and $$2008$$

C
$$2009$$ and $$2011$$

D
$$2011$$ and $$2012$$

Solution

Average production of $$2007$$ and $$2009$$ (in $$1000$$ tonnes) was
$$\dfrac{30 + 40}{2}= 35$$
Now for other given years it was
Option $$A$$ for $$2007$$ and $$2006 = \cfrac{30 + 15}{2}= 22.5$$
Option $$B$$ for $$2007$$ and $$2008 = \cfrac{30 + 20}{2}= 25$$
Option $$C$$ for $$2009$$ and $$2011 = \cfrac{40 + 30}{2}= 35$$
Option $$A$$ for $$2011$$ and $$2012 = \cfrac{30 + 60}{2}= 45$$
So it is equal to the year $$2009$$ and $$2011$$
Question 5
1 / -0

Study the bar graph and answer the question given below:
What was the percentage increase in the production of oil from $$2009$$ to $$2010$$?

A
$$10\%$$

B
$$15\%$$

C
$$20\%$$

D
$$25\%$$

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 $$= \dfrac{50- 40}{40}\times 100\%$$
$$= 25\%$$
So there was increase of $$25\%$$ from the year $$2009$$ and $$2010$$

Question 6

1 / -0

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

$$15.833$$

$$16.833$$

$$17.833$$

$$18.833$$

Solution

Answer:- By shortcut Method

Class interval width (i) = $$5-0 = 5$$

X	C. I. mid point	F	$$d=\cfrac{X-A}{i}$$	Fd
0-5	2.5	4	-1	-4
5-10	7.5=A	8	0	0
10-15	12.5	12	1	12
15-20	17.5	16	2	32
20-25	22.5	20	3	60
		$$\Sigma F = 60$$		$$\Sigma Fd=100$$

Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=7.5+\left(\cfrac{100}{60} \times 5\right)={7.5+8.33}=15.83$$

A) $$15.83$$

Question 7
1 / -0

Study the above graph and calculate the average numbers of cars manufactured by Company $$X$$ over the given period.

A
$$200{,}000$$

B
$$100{,}000$$

C
$$300{,}000$$

D
$$400{,}000$$

Solution

Average number of cars manufactured by Company $$X$$
$$=\cfrac{1}{6}[100000 + 80000 + 60000 + 120000 + 80000 + 160000]$$
$$=\cfrac{1}{6}[600000]$$
$$= 100,000$$
Question 8
1 / -0

Study the above graph and calculate the average numbers of cars manufactured by Company $$Y$$ over the given period.

A
$$96{,}666.66$$

B
$$86{,}666.66$$

C
$$76{,}666.66$$

D
$$66{,}666.66$$

Solution

Average number of cars manufactured by Company $$Y$$
$$=\cfrac{1}{6}[80000 + 60000 + 100000 + 80000 + 120000 + 140000]$$
$$= \cfrac{1}{6}[580000]$$
$$= 96{,}666.66$$
Question 9
1 / -0

Directions For Questions

Johnson's pickle factory exports bottles of pickle. The pictograph shows the number of bottles exported each day. Use the information from the graph to answer the questions.
($$1$$ pickle jar $$= 500$$)

...view full instructions

How many bottles of pickles were exported on Friday and Tuesday altogether?

A
$$4500$$

B
$$1500$$

C
$$5000$$

D
$$5500$$

Solution

$$1$$ pickle jar $$= 500$$ bottles
Friday $$= 6 \times 500 = 3000$$
Tuesday $$= 5 \times 500 = 2500$$
Altogether $$= 3000 + 2500 = 5500$$.
Question 10
1 / -0

Directions For Questions

Amit, Bala, Frank, Madesh, Thomas are officers. The pictograph shows the earnings of five officers in a month. Use the information from the graph to answer the questions.

...view full instructions

Subtracting two of the officers earnings, the difference is $$1800$$. The name of officers are

A
Amit and Bala

B
Frank and Thomas

C
Thomas and Amit

D
Amit and Madesh

Solution

Given: Subtracting two of the officers earnings, the difference is $$1800$$.
Amit $$= 6$$
Thomas $$= 2$$
So, the difference $$= 6 - 2 = 4$$
1 money bag $$= 450$$
Then, $$4 \times 450 = 1800$$
Therefore, the two officers are Thomas and Amit.

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Re-Attempt Weekly Quiz Competition

X	F	$$d=\cfrac{x-A}{i} $$	Fd
12=A	1	0	0
24	2	1	2
36	3	2	6
48	4	3	12
60	5	4	20
	$$\Sigma f = 15$$		$$\Sigma fd = 40$$

Descriptive Statistics Test 28

Result

Descriptive Statistics Test 28

Score

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Rank

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SHARING IS CARING

Question 1

$$43$$

$$44$$

$$45$$

$$46$$

Solution

Question 2

$$20000$$

$$30000$$

$$40000$$

$$50000$$

Solution

Question 3

$$100\%$$

$$200\%$$

$$300\%$$

$$400\%$$

Solution

Question 4

$$2007$$ and $$2006$$

$$2007$$ and $$2008$$

$$2009$$ and $$2011$$

$$2011$$ and $$2012$$

Solution

Question 5

$$10\%$$

$$15\%$$

$$20\%$$

$$25\%$$

Solution

Question 6

$$15.833$$

$$16.833$$

$$17.833$$

$$18.833$$

Solution

Question 7

$$200{,}000$$

$$100{,}000$$

$$300{,}000$$

$$400{,}000$$

Solution

Question 8

$$96{,}666.66$$

$$86{,}666.66$$

$$76{,}666.66$$

$$66{,}666.66$$

Solution

Question 9

$$4500$$

$$1500$$

$$5000$$

$$5500$$

Solution

Question 10

Amit and Bala

Frank and Thomas

Thomas and Amit

Amit and Madesh

Solution

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