Statistics Test

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Statistics Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

If in a frequency distribution, the mean and median are $$21$$ and $$22$$ respectively, then its mode is approximately

A
$$22.0$$

B
$$20.5$$

C
$$25.5$$

D
$$24.0$$

Solution

We have,
$$Mode + 2 \times Mean =3\times Median$$
$$\Rightarrow Mode = 3 \times Median - 2\times Mean$$
Given : $$Mean=21$$ and $$Median =22$$
Putting the given values, we get
$$Mode =3\times 22-2\times 21=66-42=24$$.
Hence, option D is correct.
Question 2
1 / -0

Find mode from the following grouped frequency distribution table:
Number of Trees Planted Number of Students
$$0-10$$ $$30$$
$$10-20$$ $$60$$
$$20-30$$ $$50$$
$$30-40$$ $$70$$
$$40-50$$ $$40$$

A
$$38$$ trees

B
$$36$$ trees

C
$$35$$ trees

D
$$34$$ trees

Solution

$$N=30+60+50+70+40=250$$
$$\implies \cfrac{N}{2}=125$$
Since, the highest frequency of students is $$70$$. So, $$30-40$$ is the modal group.
$$\implies L=30$$
Also, $$f_{0}=50$$, frequency of the group before the modal group
$$f_{1}=70$$, frequency of the modal group
$$f_{2}=40$$, frequency of the group after modal group.
$$h=10$$, width of the group
Then, Mode $$=L + \cfrac{f_1 - f_0}{2f_1 - f_0-f_2}\times h$$
$$=30+\cfrac{70-50}{140-50-40}\times 10$$
$$=30+\cfrac{20}{50}\times 10$$
$$=30+4=34$$
Hence, mode is $$34$$ trees.
Question 3
1 / -0

Class 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35
Frequency 5 15 6 10 14 9
The lower limit of the modal class of the following data is :

A
25

B
15

C
10

D
30

Solution

The highest frequency occurs in the class range $$10-15$$ that is $$15$$.
Therefore,
Modal Class: $$10-15$$
Lower Limit: $$10$$
Question 4
1 / -0

Which of the following is a measure of central tendency?

A
Mean

B
Median

C
Mode

D
All of the above

Solution

A measure of central tendency is a typical value that is referred to as a representative of the whole data.
The most common measures of central tendency are mean median and mode.
Question 5
1 / -0

Median of $$15, 28, 72, 56, 44, 32, 31, 43\ and\ 51\ is\ 43.$$

A
True

B
False

C
Neither

D
Either

Solution

The terms are: 15, 28, 72, 56, 44, 32, 31, 43 and 51.
Arranging them in ascending order: 15, 28, 31, 32, 43, 44, 51, 56, 72

Since the total number of terms is odd that is 9, therefore the median will be the middle term that is the 5th term which is 43.
Question 6
1 / -0

A child says that the median of $$3, 14, 18, 20, 5$$ is $$18$$. What concept does the child missed about finding the median?

A
The order of numbers.

B
$$14$$

C
$$18$$

D
None of these

Solution

To calculate the median of any data series. The data series has to be arranged in the ascending order. The child hasn't arranged the data series in ascending order.
Question 7
1 / -0

State true or false:
The mode is the most frequently occurring observation.

A
True

B
False

C
Can't determine

D
None of these

Solution

The observation occurring the most number of times or which has highest frequency is called the mode.
Thus, the given statement is true.
Question 8
1 / -0

3 median $$=$$ mode + ................. mean.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

A distribution in which the values of mean, median and mode coincide (i.e. mean = median = mode) is known as a symmetrical distribution. Conversely, when values of mean, median and mode are not equal the distribution is known as asymmetrical or skewed distribution. In moderately skewed or asymmetrical distribution a very important relationship exists among these three measures of central tendency. In such distributions the distance between the mean and median is about one-third of the distance between the mean and mode, as will be clear from the diagrams 1 and 2. Karl Pearson expressed this relationship as:
Mode = mean - 3 [mean - median]
Mode = 3 median - 2 mean
3 Median = Mode + 2 Mean
Question 9
1 / -0

The modal value is the value of the variate which divides the total frequency into two equal parts.

A
True

B
False

C
Neither

D
Either

Solution

False. Modal value is the value which occurs maximum number of times in the data.
Question 10
1 / -0

Median divides the total frequency into _____ equal parts.

A
$$2$$

B
$$3$$

C
$$4$$

D
None of these

Solution

The median of the data series is the middle term or the mean of the two middle terms.
Hence, it divides the data series or the frequency of terms into two equal halves.

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Re-Attempt Weekly Quiz Competition

Number of Trees Planted	Number of Students
$$0-10$$	$$30$$
$$10-20$$	$$60$$
$$20-30$$	$$50$$
$$30-40$$	$$70$$
$$40-50$$	$$40$$

Class	5 - 10	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35
Frequency	5	15	6	10	14	9

Statistics Test - 14

Result

Statistics Test - 14

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SHARING IS CARING

Question 1

$$22.0$$

$$20.5$$

$$25.5$$

$$24.0$$

Solution

Question 2

$$38$$ trees

$$36$$ trees

$$35$$ trees

$$34$$ trees

Solution

Question 3

25

15

10

30

Solution

Question 4

Mean

Median

Mode

All of the above

Solution

Question 5

True

False

Neither

Either

Solution

Question 6

The order of numbers.

$$14$$

$$18$$

None of these

Solution

Question 7

True

False

Can't determine

None of these

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 9

True

False

Neither

Either

Solution

Question 10

$$2$$

$$3$$

$$4$$

None of these

Solution

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