Measures of Dispersion Test

Result

Measures of Dispersion Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

The mean of distribution is $$22.2$$ and its mode is $$23.3$$. The median is _______.

A
$$26.5$$

B
$$22.6$$

C
$$28.5$$

D
$$26.6$$
Question 2
1 / -0

If the group data has open end classes, one cannot calculate ________.

A
median

B
mode

C
mean

D
quartiles
Question 3
1 / -0

If modal value is not clear in a distribution, it can be ascertained by the method of ________.

A
grouping

B
guessing

C
summarizing

D
trial and error
Question 4
1 / -0

If for a series the arithmetic mean is $$25$$ and the harmonic mean is $$9$$, what is the geometric mean?

A
$$25$$

B
$$15$$

C
$$28$$

D
$$16$$
Question 5
1 / -0

The standard deviation of $$10, 16, 10, 16, 10, 10, 16, 16$$ is?

A
$$4$$

B
$$6$$

C
$$3$$

D
$$0$$

Solution

Assume $$A=13;$$ hence $$d=A-13$$
x d $$d^2$$
$$10$$ $$-3$$ $$9$$
$$16$$ $$3$$ $$9$$
$$10$$ $$-3$$ $$9$$
$$16$$ $$3$$ $$9$$
$$10$$ $$-3$$ $$9$$
$$10$$ $$-3$$ $$9$$
$$16$$ $$-3$$ $$9$$
$$16$$ $$3$$ $$9$$

$$\displaystyle\sum d=0$$ $$\displaystyle\sum d^2=72$$
Use following formula:
$$\sqrt{\displaystyle\frac{\displaystyle\sum d^2}{n}-\left[\displaystyle\frac{\displaystyle\sum d}{n}\right]^2}$$
$$\sqrt{\displaystyle\frac{72}{8}-\left[\displaystyle\frac{0}{8}\right]^2}$$
$$\sqrt{9}$$
$$=3$$.
Question 6
1 / -0

Directions For Questions

The following table gives the distribution of daily wages of $$900$$ workers. However, the frequencies of the two classes $$40-50$$ and $$60-70$$ are missing. If the median of the distribution is Rs. $$59.25$$, find the missing figures.

...view full instructions

Class Intervals $$30-40$$ $$40-50$$ $$50-60$$ $$60-70$$ $$70-80$$
Frequency $$120$$ ? $$200$$ ? $$185$$
Cumulative Frequency $$120$$ ? ? ? $$900$$
Cumulative frequency for class $$50-60=$$?

A
$$715$$

B
$$465$$

C
$$145$$

D
$$265$$

Solution

Class Intervals Wages Frequency(f) Cumulative Frequency
$$30-40$$ $$120$$ $$120$$
$$40-50$$ $$f_1$$ $$120+f_1$$
$$50-60$$ $$200$$ $$320+f_1$$
$$60-70$$ $$f_2$$ $$320+f_1+f_2$$
$$70-80$$ $$185$$ $$900$$
Since median is given to be $$59.25$$, the median class is $$50-60$$. Thus, we can write, solving this for $$f_1$$, we get, $$f_1=145$$; Further, $$f_2=900-(120+145+200+185)=250$$.
Question 7
1 / -0

The geometric mean of $$3, 6, 24 and 48$$ is ______.

A
$$32$$

B
$$12$$

C
$$14$$

D
$$24$$
Question 8
1 / -0

Sum of the deviations about mean is ______.

A
zero

B
minimum

C
maximum

D
one
Question 9
1 / -0

Harmonic mean gives more weight age to _______.

A
small values

B
large values

C
positive values

D
negative values
Question 10
1 / -0

In a frequency distribution with open ends, one cannot find out ______.

A
mean

B
median

C
mode

D
all of the above

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x	d	$$d^2$$
$$10$$	$$-3$$	$$9$$
$$16$$	$$3$$	$$9$$
$$10$$	$$-3$$	$$9$$
$$16$$	$$3$$	$$9$$
$$10$$	$$-3$$	$$9$$
$$10$$	$$-3$$	$$9$$
$$16$$	$$-3$$	$$9$$
$$16$$	$$3$$	$$9$$
	$$\displaystyle\sum d=0$$	$$\displaystyle\sum d^2=72$$

Class Intervals	$$30-40$$	$$40-50$$	$$50-60$$	$$60-70$$	$$70-80$$
Frequency	$$120$$	?	$$200$$	?	$$185$$
Cumulative Frequency	$$120$$	?	?	?	$$900$$

Class Intervals Wages	Frequency(f)	Cumulative Frequency
$$30-40$$	$$120$$	$$120$$
$$40-50$$	$$f_1$$	$$120+f_1$$
$$50-60$$	$$200$$	$$320+f_1$$
$$60-70$$	$$f_2$$	$$320+f_1+f_2$$
$$70-80$$	$$185$$	$$900$$

Measures of Dispersion Test - 14

Result

Measures of Dispersion Test - 14

Score

-

Rank

-

SHARING IS CARING

Question 1

$$26.5$$

$$22.6$$

$$28.5$$

$$26.6$$

Question 2

median

mode

mean

quartiles

Question 3

grouping

guessing

summarizing

trial and error

Question 4

$$25$$

$$15$$

$$28$$

$$16$$

Question 5

$$4$$

$$6$$

$$3$$

$$0$$

Solution

Question 6

$$715$$

$$465$$

$$145$$

$$265$$

Solution

Question 7

$$32$$

$$12$$

$$14$$

$$24$$

Question 8

zero

minimum

maximum

one

Question 9

small values

large values

positive values

negative values

Question 10

mean

median

mode

all of the above

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