We have the following table:

(i) (c): Here, it is given that total frequency = 100
\(\therefore\) 76 + x + y = 100 \(\Rightarrow\) x + y = 24
(ii) (c): Here \(\frac{N}{2}=\frac{100}{2}=50\)
Also, median = 525
\(\therefore\) Median class is 500-600.
\(\text { Now, median }=l+\left(\frac{N / 2-c . f .}{f}\right) \times h \)
\(\Rightarrow 525=500+\left(\frac{50-(36+x)}{20}\right) \times 100 \)
\(\Rightarrow 5=50-36-x \Rightarrow x=9\)
(iii) (b) : Since, maximum frequency is 20, so modal class is 500 - 600. Hence, upper limit of modal class is 600.
(iv) (b) : Since, x + y = 24 \(\Rightarrow\) y = 24 - 9 = 15
Required average consumption
\(\begin{aligned} & 50 \times 2+150 \times 5+250 \times 9+350 \times 12+450 \times 17 \\ =& \frac{+550 \times 20+650 \times 15+750 \times 9+850 \times 7+950 \times 4}{100} \\ =& \frac{52200}{100}=522 \mathrm{kwh} \end{aligned}\)
(v) (d)

(i) (b): The age of any person is a positive number, so the first class must be 0 - 10.
(ii) (c):
Let us consider the following table:

Here, N = 71, therefore \(\frac{N}{2}=35.5\)
Now, the class interval whose cumulative frequency is
just greater than 35.5 is 30-40.
\(\therefore\) Median class = 30-40
(iii) (a): Clearly, the cumulative frequency of the class
preceding the median class is 22.
(iv) (d): Median \(=l+\left(\frac{\frac{N}{2}-\varsigma . f .}{f}\right) \times h\)
\(=30+\left(\frac{35.5-22}{18}\right) \times 10=30+13.5 \times \frac{10}{18}=30+7.5=37.5\)
Thus, the median age of the persons visited the
museum is 37.5 years
(v) (c): Number of persons, whose age lying in 30-40 = 18
\(\therefore\) Total amount spent by people of this group
= Rs (30 x 18) = Rs 540

(i) (b): The upper limit of a class and the lower class of its succeeding class differ by 1.
(ii) (d) : Here, class intervals are in inclusive form. So, we first convert them in exclusive form. The frequency distribution table in exclusive form is as follows:

\(\text { Here, } \Sigma f_{i} \text { i.e., } N=60 \)
\(\Rightarrow \frac{N}{2}=30\)

Now, the class interval whose cumulative frequency is
just greater than 30 is 219.5 - 229.5.
\(\therefore\) Median class is 219.5 - 229.5.
(iii) (b): Clearly, the cumulative frequency of the class preceding the median class is 18
(iv) (d): Median \(=l+\left[\frac{\frac{N}{2}-c . f .}{f}\right] \times h\)
\(=219.5+\left(\frac{30-18}{26}\right) \times 10 \)
\(=219.5+\frac{12 \times 10}{26}=219.5+4.62=224.12\)
\(\therefore\) Median of the distance travelled is 224.12 km
(v) (d): We know, Mode = 3 Median - 2 Mean
\(\therefore \quad \text { Mean }=\frac{1}{2}(3 \text { Median }-\text { Mode }) \)
\(=\frac{1}{2}(672.36-223.78)=224.29 \mathrm{~km}\)

Let us consider the following table:

(i) (d): Clearly, the possible values of assumed mean (A) are 35, 45, 55, 65, 75.
(ii) (c): The values of |d_i| are 0, 10,20
Thus, the minimum value of |d_i| is 0.
(iii) (b): Required Mean \(=A+\frac{\sum f_{i} d_{i}}{\sum f_{i}}=55-\frac{1200}{420}\)
= Rs 52.14.
(iv) (a): Mean by direct and assumed mean method are always equal.
(v) (d): Average toll tax received by a vehicle = Rs 52.14 Total number of vehicles = 420
\(\therefore\) Average toll tax received in a day = Rs (52.14 x 420) = Rs 21898.80

(i) (c): Class mark of class 25 - 29
\(=\frac{25+29}{2}=\frac{54}{2}=27\)
(ii) (d): Let us consider the following table:

\(\therefore \quad \operatorname{Mean}(\bar{x})=\frac{\sum x_{i} f_{i}}{\sum f_{i}}=\frac{1265}{50}=25.3\)

Thus, the mean time to complete the work for a worker
= 25.3 hrs = 3 days
(iii) (d)
(iv) (a)
(v) (d): We know the measure of central tendency are mean, median and mode.