(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

(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.

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): 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
\(\therefore \quad \text { Mean }=\frac{1}{2}(3 \text { Median }-\text { Mode }) \)
\(=\frac{1}{2}(672.36-223.78)=224.29 \mathrm{~km}\)

(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

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.

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) (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.

(i) (d): We know that,
\(\text { Class mark }=\frac{\text { Lower limit }+\text { Upper limit }}{2} \)
\(\Rightarrow m=\frac{\text { Lower limit }+b}{2} \Rightarrow \text { Lower limit }=2 m-b\)
(ii) (a):

\(\begin{aligned} &\therefore \quad \text { Average lifetime of a packet }\\ &=A+\frac{\sum f_{i} d_{i}}{\sum f_{i}}=325+\frac{6400}{400}=341 \mathrm{hrs} \end{aligned}\)

(iii) (b) : \(\text { Here, } N=400 \Rightarrow \frac{N}{2}=200\)
Also, cumulative frequency for the given distribution are 14, 70, 130,216,290,352,400
\(\therefore\) c.f just greater than 200 is 216, which is
corresponding to the interval 300-350.
l= 300, f=86, c.f = 130, h = 50
\(\therefore \quad \text { Median }=l+\left(\frac{\frac{N}{2}-c . f .}{f}\right) \times h=300+\left(\frac{200-130}{86}\right) \times 50\)
= 300 + 40.697 = 340.697 ""340 hrs (approx.)
(iv) (a) : We know that Mode = 3 Median - 2 Mean
= 3(340.697) -2(341)
= 1022.091 - 682 = 340.091 ""340 hrs
(v) (c): Since, minimum of mean, median and mode is approximately 340 hrs. So, manufacturer should claim that lifetime of a packet is 340 hrs.

(i) (a): If f_i and x_i are very small, then direct method is appropriate method for calculating mean.
(ii) (a) : The frequency distribution table from the given data can be drawn as :

\(\therefore \quad \text { Mean }=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{326}{40}=8.15 \text { litres }\)

(iii) (b) : If 2000 vehicles comes daily and average quantity of diesel required for a vehicle is 8.15 litres, then total quantity of diesel required = 2000 x 8.15
= 16300 litres
(iv) (c) : Here \(N=40 \text { and } \frac{N}{2}=20\) c.f for the distribution are 5, 15,25,32,40
Now, c.f just greater than 20 is 25 which is corresponding to the class interval 7-9.
So median class is 7-9.
\(\therefore\) Required sum of upper limit and lower limit
= 7 + 9 = 16
(v) (b): We know, Mode = 3 Median -2 Mean
= 3(8) - 2(8.15) = 24 - 16.3 = 7.7

(i) (a): \(n=51 . \mathrm{So}, \frac{n}{2}=\frac{51}{2}=25.5 .\)
This observation lies in the class 145 - 150
= 145
(ii) (b): 150
(iii) (a): 145
(iv) (c): 5
(v) (a): l ( lower limit) = 145, cf = 11, f = 18, h = 5.
\(\text { Median }=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) \times h\)
\(\text { Median }=145+\left(\frac{25.5-11}{18}\right) \times 5=145+\frac{72.5}{18}\)
= 149.03 cm

(i) (b): Here N =35
\(\therefore \frac{N}{2}=\frac{35}{2}=17.5\)
Since, the cumulative frequency is greater than 17.5 is 28 and the corresponding class intervals is 46 - 48.
Median class = 46 - 48
(ii) (a): Here, l =46, f = 14 cf = 14 and h = 2
\(\therefore \text { Median }=l+\left(\frac{\frac{N}{2}-c f}{f}\right) \times h\)
\(=46+\frac{17.5-14}{14} \times 2\)
\(=46+\frac{3.5}{14} \times 2=46+0.5=46.5\)
(iii) (b): \(\text { Mean, } \bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{1603}{35}=45.8\)
(iv) (b): The highest frequency in the given data is 14, whose modal class is 46-48.
(v) (b): While computing the mean of grouped data, we assume that the frequencies are centered at the class marks of the classes.

(i) (a): 

Here, \(\Sigma f_{i}=n=68 \text { then } \frac{n}{2}=\frac{68}{2}=34\) which lies in interval 125 - 145
= 125
(ii) (a): 125
(iii) (b): 145
(iv) (c): 20
(v) (a): Median class = 125 - 145
So, l = 125; n = 68; f = 20; cf = 22 and h = 20
Using formula, Median \(=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h\)
\(=125+\left\{\frac{\frac{68}{2}-22}{20}\right\} \times 20\)
\(=125+\frac{34-22}{20} \times 20=125+12\)
= 137

(i) (c): 

\(\text { Mean }=\frac{1720}{40}\)
= 43
(ii) (c): 60
(iii) (b): Median
(iv) (c): Median class 40 - 60, Modal class = 40 - 60
Sum of lower limits of median class and modal class = 40 + 40 = 80
(v) (c): Number of students are = 8 + 10 + 13
= 31

(a) (ii) Since the highest frequency is 23 which belongs to 35 – 45.
Therefore, modal class is 35 – 45.
(b) (ii) Here, \(n=80 \Rightarrow \frac{n}{2}=40\) 
which lies in 35 – 45 
Therefore, medial class is 35 – 45.
(c) (iii)   Here,l = 35, f₀ = 21, f₁= 23, f₂ = 14, h = 10.
\(\text { Mode }=l+\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \times h \quad \Rightarrow \text { Mode }=35+\frac{23-21}{46-21-14} \times 10\) 
(d) (i) 35-45
(e) (ii)19