Question

Write the correct alternative answer for each of the following questions.

(i) Which of the following data is not primary ?

(A) By visiting a certain class, gathering information about attendence of students.

(B) By actual visit to homes, to find number of family members.

(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

(D) Review the cleanliness status of canals by actually visitiing them.

(ii) What is the upper class limit for the class 25-35 ?

(A) 25

(B) 35

(C) 60

(D) 30

(iii) What is the class-mark of class 25-35 ?

(A) 25

(B) 35

(C) 60

(D) 30

(iv) If the classes are 0-10, 10-20, 20-30... then in which class should the observation 10 be included ?

(A) 0-10 (B) 10-20 (C) 0-10 and 10-20 in these 2 classes (D) 20-30

(v) If $\overline{x}$ is the mean of $x_{1,}{x}_2.............x_n$and $\overline{y}$ is the mean of $y_1,y_2................y_n$ and $\overline{z}$ is the mean $x_1,x_2..............x_n,y_1,y_2,.....................y_n$ then z = ?

(A) $\frac{\overline{x}+\overline{y}}{2}$ ( B) $\overline{x}+\overline{y}$ (C) $\frac{\overline{x}+\overline{y}}{n}$ (D) $\frac{\overline{x}+\overline{y}}{2n}$

(vi) The mean of five numbers is 80, out of which mean of 4 numbers is 46, find the 5th number :

(A) 4

(B) 20

(C) 434

(D) 66

(vii) Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.

(A) 40.6

(B) 40.4

(C) 40.3

(D) 40.7

(viii) What is the the mode of 19, 19, 15, 20, 25, 15, 20, 15?

(A) 15,

(B) 20

(C) 19
(D) 25

(ix) What is the median of 7, 10, 7, 5, 9, 10 ?

(A) 7

(B) 9

(C) 8

(D) 10

(x) From following table, what is the cumulative frequency of less than type for the class 30-40 ?

Class	0-10	10-20	20-30	30-40	40-50
Frequency	7	3	12	13	2

(A) 13

(B) 15

(C) 35

(D) 22

Answer 1

(i)

(A) By visiting a certain class, gathering information about attendence of students.

Since, the information related to the attendence has been already recorded.

So, it is not primary.

(B) By actual visit to homes, to find number of family members.

Since, the information is collected by visiting.

So, it is a primary data.

(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.

Since, the information is collected by each member.

So, it is a primary data.

(D) Review the cleanliness status of canals by actually visitiing them.

Since, this information is gathered by visiting.

So, it is a primary data.

Hence, the correct alternative answer is (A).

(ii) Since, the upper class limit for the class 25-35 is 35.

Hence, the correct alternative answer is (B).

(iii) Since, the class-mark of class 25-35 $\frac{25+35}{2}=\frac{60}{2}=30$

Hence, the correct alternative answer is (D).

(iv) If the classes are 0-10, 10-20, 20-30..,. then the class in which the observation 10 should be included 10-20.

Hence, the correct alternative answer is (B).

(v) If $\overline{x}$ is the mean of $x_{1,}{x}_2.............x_n$and $\overline{y}$ is the mean of $y_1,y_2................y_n$ and $\overline{z}$ is the mean $x_1,x_2..............x_n,y_1,y_2,.....................y_n$,

Then z = $\frac{n\overline{x}+n\overline{y}}{2n}=\frac{{\displaystyle n\left(\overline{x}+\overline{y}\right)}}{2n}=\frac{{\displaystyle \overline{x}+\overline{y}}}{2}$

Hence, the correct alternative answer is (A).

(vi) Since, the mean of five numbers is 80, out of which mean of 4 numbers is 46

So, the 5th number $=\mathrm{Sum}\mathrm{of}5\mathrm{observations}-\mathrm{Sum}\mathrm{of}4\mathrm{observations}=\left(80\times 5\right)-\left(46\times 4\right)=400-184=216$.

Disclaimer: The options provided are all incorrect.

(vii) We have,

Mean of 100 observations is 40.

Total sum of observations = $100\times 40=4000$

If the 9th observation is 30 and is replaced by 70 keeping all other observations same.

New, total sum of observations = $4000-30+70=4040$

So, the new mean = $\frac{4040}{100}=40.4$

Hence, the correct alternative answer is (B).

(viii) Since, the observation 15 is occuring the most in the data: 19, 19, 15, 20, 25, 15, 20, 15

So, the mode of the given data is 15.

Hence, the correct alternative answer is (A).

(ix) Since, the ascending order of the given data is as: 5, 7, 7, 9, 10, 10.

And, the number of observations is 6, which is even.

$$\begin{gathered} \mathrm{So},\mathrm{the}\mathrm{median}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{data}=\frac{1}{2}\left[\left(\frac{6}{2}\right)\mathrm{th}\mathrm{observation}+\left(\frac{6}{2}+1\right)\mathrm{th}\mathrm{observation}\right] \\ =\frac{1}{2}\left[3\mathrm{rd}\mathrm{observation}+4\mathrm{th}\mathrm{observation}\right] \\ =\frac{1}{2}\left[7+9\right] \\ =\frac{16}{2} \\ =8 \end{gathered}$$

Hence, the correct alternative answer is (C).

(x) We have,

Class	0-10	10-20	20-30	30-40	40-50
Frequency	7	3	12	13	2

So, the cumulative frequency of less than type for the class 30-40 = $7+3+12+13=35$

Hence, the correct alternative answer is (C).