(i) (d): Required number of persons = 9 + 6 = 15
(ii) (c): Required number of persons = 6 + 8 + 2 = 16
(iii) (a) : 50-60 is the modal class as the maximum frequency is 9.
(iv) (b) : The cumulative frequency distribution table for the given data can be drawn as :

\(\text { Here, } \frac{N}{2}=\frac{25}{2}=12.5\)
The cumulative frequency just greater than 12.5 lies in the interval 60-70.
Hence, the median class is 60-70.
(v) (a): We know, Mode = 3 Median - 2 Mean
\(\therefore\) 3 Median = Mode + 2 Mean.

Given frequency distribution table can be drawn as:

(i) (d): Clearly, average mileage
\(=\frac{7240}{50}=144.8 \mathrm{~km} / \text { charge }\)
(ii) (b) : Since, highest frequency is IS, therefore,
modal class is 140-160.
Here, l = 140,f₁ = 18,f₀ = 12,f₂ = 13, h = 20
\(\therefore \quad \text { Mode }=140+\frac{18-12}{36-12-13} \times 20=140+\frac{6}{11} \times 20 \)
\(=140+\frac{120}{11}=140+10.91=150.91\)
(iii) (b) : Here \(\frac{N}{2}=\frac{50}{2}=25\) and the corresponding class whose cumulative frequency is just greater than
25 is 140-160.
Here, l = 140, c.f = 19, h = 20 and f= 18
\(\therefore \quad \text { Median }=l+\left(\frac{\frac{N}{2}-c . f .}{f}\right) \times h\)
\(=140+\frac{25-19}{18} \times 20=140+\frac{60}{9}=146.67\)
(iv) (a) : Assumed mean method is useful in determining the mean.
(v) (a): Since, Mean = 144.S, Mode = 150.91 and Median = 146.67 and minimum of which is 144 approx, therefore manufacturer can claim the mileage for his scooter 144 km/charge.

(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) (b): Since, the highest frequency is 70, therefore the maximum number of children are of the age-group 10-12.
(ii) (a): Since, the modal class is 10-12
\(\therefore\) Lower limit of modal class = 10
(iii) (c) : Here,f₀ = 58,f₁ = 70 and f₂ = 42
Thus, the frequency of the class succeeding the modal class is 42.
(iv) (d): Mode \(=l+\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right] \times h\)
\(=10+\left[\frac{70-58}{140-58-42}\right] \times 2\)
\(=10+\frac{12}{40} \times 2=10+\frac{24}{40}=10.6 \text { years }\)
(v) (c): Given that, Mean = Mode
\(\therefore\) By Empirical relation, we have
Mode = 3 Median - 2 Mean
\(\Rightarrow\) Mode = 3 Median - 2 Mode
\(\Rightarrow\) 3 Mode = 3 Median
\(\Rightarrow\) Median = Mode = Mean