EXERCISE 13.1

1. Identify class intervals and frequencies
\[\begin{array}{c|c}
\text{Class (plants)} & \text{Number of houses (}f\text{)}\\ \hline
0-2 & 1\\
2-4 & 2\\
4-6 & 1\\
6-8 & 5\\
8-10 & 6\\
10-12 & 2\\
12-14 & 3\\
\end{array}\]
Total \(\sum f = 20\).

2. Find the class‑midpoint (assumed value) for each interval
\[\begin{array}{c|c}
\text{Class} & \text{Midpoint }(x)\\ \hline
0-2 & \frac{0+2}{2}=1\\
2-4 & \frac{2+4}{2}=3\\
4-6 & \frac{4+6}{2}=5\\
6-8 & \frac{6+8}{2}=7\\
8-10 & \frac{8+10}{2}=9\\
10-12 & \frac{10+12}{2}=11\\
12-14 & \frac{12+14}{2}=13\\
\end{array}\]

3. Compute \(f\times x\) for each class
\[\begin{array}{c|c|c}
\text{Midpoint }(x) & f & f\times x\\ \hline
1 & 1 & 1\\
3 & 2 & 6\\
5 & 1 & 5\\
7 & 5 & 35\\
9 & 6 & 54\\
11 & 2 & 22\\
13 & 3 & 39\\
\end{array}\]
Sum \(\sum f x = 1+6+5+35+54+22+39 = 162\).

4. Apply the formula for mean
\[\bar{x}=\frac{\sum f x}{\sum f}=\frac{162}{20}=8.1\]

5. Interpretation
The average (mean) number of plants per house in the surveyed locality is 8.1 plants.

6. Method used and justification
- *Method*: Assumed‑mean (grouped data) method using class midpoints.
- *Why*: The original data are given in class intervals, not as individual observations. By taking the midpoint of each interval as a representative value, we can compute an approximate mean for the grouped data, which is the standard technique prescribed in the NCERT textbook for such problems.

1. Form the class‑midpoints (also called class‑marks).
\[\begin{array}{c|c}
\text{Class interval (₹)} & \text{Midpoint } x_i \\ \hline
500-520 & \frac{500+520}{2}=510 \\
520-540 & \frac{520+540}{2}=530 \\
540-560 & \frac{540+560}{2}=550 \\
560-580 & \frac{560+580}{2}=570 \\
580-600 & \frac{580+600}{2}=590 \\
\end{array}\]

2. Multiply each midpoint by its frequency $f_i$.
\[\begin{array}{c|c|c}
x_i & f_i & f_i x_i \\ \hline
510 & 12 & 12\times510 = 6120 \\
530 & 14 & 14\times530 = 7420 \\
550 & 8 & 8\times550 = 4400 \\
570 & 6 & 6\times570 = 3420 \\
590 & 10 & 10\times590 = 5900 \\ \hline
\text{Total} & \sum f_i = 50 & \sum f_i x_i = 27260 \\
\end{array}\]

3. Apply the formula for the mean.
$$\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{27260}{50}=545.2$$

4. Interpretation: The average (mean) daily wage of the 50 workers is ₹ 545.2.

1. Find the class mid‑points (average of lower and upper limits):
\[
\begin{aligned}
11-13 &: \; x_1 = \frac{11+13}{2}=12 \\
13-15 &: \; x_2 = \frac{13+15}{2}=14 \\
15-17 &: \; x_3 = \frac{15+17}{2}=16 \\
17-19 &: \; x_4 = \frac{17+19}{2}=18 \\
19-21 &: \; x_5 = \frac{19+21}{2}=20 \\
21-23 &: \; x_6 = \frac{21+23}{2}=22 \\
23-25 &: \; x_7 = \frac{23+25}{2}=24
\end{aligned}
\]

2. Write the expressions for \(\sum f_i\) and \(\sum f_i x_i\)
\[
\sum f_i = 7+6+9+13+f+5+4 = 44+f
\]
\[
\sum f_i x_i = 12\times7 + 14\times6 + 16\times9 + 18\times13 + 20\times f + 22\times5 + 24\times4
\]
Calculating the known products:
\[
\begin{aligned}
12\times7 &= 84 \\
14\times6 &= 84 \\
16\times9 &= 144 \\
18\times13 &= 234 \\
22\times5 &= 110 \\
24\times4 &= 96
\end{aligned}
\]
Hence
\[
\sum f_i x_i = 84+84+144+234+20f+110+96 = 752 + 20f
\]

3. Use the given mean (\(\bar{x}=18\))
\[
\bar{x}=\frac{\sum f_i x_i}{\sum f_i}\;\Rightarrow\; 18 = \frac{752+20f}{44+f}
\]

4. Solve the linear equation for \(f\)
\[
752 + 20f = 18(44+f) = 792 + 18f
\]
\[
20f - 18f = 792 - 752 \quad\Rightarrow\quad 2f = 40
\]
\[
f = \frac{40}{2} = 20
\]

5. Check (optional)
Total frequency = 44 + 20 = 64.
\(\displaystyle \frac{752+20\times20}{64}=\frac{1152}{64}=18\), confirming the mean.

Therefore, the missing frequency is \(f = 20\).

1. Identify the class‑marks (mid‑points) for each class:\\
\[\begin{aligned}
65-68 &: \frac{65+68}{2}=66.5 \\
68-71 &: \frac{68+71}{2}=69.5 \\
71-74 &: \frac{71+74}{2}=72.5 \\
74-77 &: \frac{74+77}{2}=75.5 \\
77-80 &: \frac{77+80}{2}=78.5 \\
80-83 &: \frac{80+83}{2}=81.5 \\
83-86 &: \frac{83+86}{2}=84.5
\end{aligned}\]
2. Multiply each class‑mark by its frequency (\(f_i x_i\)) and tabulate:\\
\[\begin{array}{c|c|c}
\text{Class} & f_i & f_i x_i \\ \hline
65-68 & 2 & 2\times66.5 = 133 \\
68-71 & 4 & 4\times69.5 = 278 \\
71-74 & 3 & 3\times72.5 = 217.5 \\
74-77 & 8 & 8\times75.5 = 604 \\
77-80 & 7 & 7\times78.5 = 549.5 \\
80-83 & 4 & 4\times81.5 = 326 \\
83-86 & 2 & 2\times84.5 = 169 \\ \hline
\text{Total} & \sum f_i = 30 & \sum f_i x_i = 2277
\end{array}\]
3. Apply the mean formula:\\
\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2277}{30}=75.9\]
4. Interpretation: The average (mean) heart‑beat rate of the 30 women is 75.9 beats per minute.

1. Form the class‑mark (mid‑point) for each interval
\[\begin{aligned}
50-52 &: x_1 = \frac{50+52}{2}=51 \\
53-55 &: x_2 = \frac{53+55}{2}=54 \\
56-58 &: x_3 = \frac{56+58}{2}=57 \\
59-61 &: x_4 = \frac{59+61}{2}=60 \\
62-64 &: x_5 = \frac{62+64}{2}=63
\end{aligned}\]
2. Write the frequencies
\[f_1=15,\; f_2=110,\; f_3=135,\; f_4=115,\; f_5=25\]
3. Compute \(f_i x_i\) for each class
\[\begin{aligned}
f_1x_1 &= 15\times51 = 765 \\
f_2x_2 &= 110\times54 = 5940 \\
f_3x_3 &= 135\times57 = 7695 \\
f_4x_4 &= 115\times60 = 6900 \\
f_5x_5 &= 25\times63 = 1575
\end{aligned}\]
4. Find the totals
\[\sum f_i = 15+110+135+115+25 = 400\]
\[\sum f_i x_i = 765+5940+7695+6900+1575 = 22875\]
5. Apply the mean formula
\[\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{22875}{400}=57.1875\]
6. State the answer (to two decimal places)
\[\boxed{\bar{x}\approx 57.19\text{ mangoes per box}}\]
7. Method chosen: The "class‑mark (mid‑point) method" for grouped data was used, which is the standard technique taught in NCERT for finding the mean of a frequency distribution.

1. Identify the class intervals and their frequencies
\[\begin{array}{c|c}
\text{Class interval (₹)} & \text{Frequency (f)}\\ \hline
100-150 & 4\\
150-200 & 5\\
200-250 & 12\\
250-300 & 2\\
300-350 & 2\\
\end{array}\]
Total frequency \(N = 4+5+12+2+2 = 25\).

2. Find the class‑midpoint (x) for each interval
\[x = \frac{\text{lower limit}+\text{upper limit}}{2}\]
\[\begin{array}{c|c}
\text{Class interval} & \text{Midpoint (x)}\\ \hline
100-150 & 125\\
150-200 & 175\\
200-250 & 225\\
250-300 & 275\\
300-350 & 325\\
\end{array}\]

3. Compute \(f\times x\) for each class
\[\begin{array}{c|c|c}
\text{Midpoint (x)} & \text{Frequency (f)} & f\times x\\ \hline
125 & 4 & 500\\
175 & 5 & 875\\
225 & 12 & 2700\\
275 & 2 & 550\\
325 & 2 & 650\\
\end{array}\]
Sum of \(f\times x\): \(\sum f x = 500+875+2700+550+650 = 5275\).

4. Calculate the mean
\[\bar{x} = \frac{\sum f x}{N} = \frac{5275}{25} = 211\]

5. Interpretation
The average (mean) daily expenditure on food per household is \(\mathbf{₹\,211}\).

1. List the classes, frequencies and find the class‑midpoints
\[
\begin{array}{c|c|c}
\text{Class (ppm)} & f_i & x_i \text{ (midpoint)}\\ \hline
0.00-0.04 & 4 & \frac{0.00+0.04}{2}=0.02\\
0.04-0.08 & 9 & \frac{0.04+0.08}{2}=0.06\\
0.08-0.12 & 9 & \frac{0.08+0.12}{2}=0.10\\
0.12-0.16 & 2 & \frac{0.12+0.16}{2}=0.14\\
0.16-0.20 & 4 & \frac{0.16+0.20}{2}=0.18\\
0.20-0.24 & 2 & \frac{0.20+0.24}{2}=0.22\\
\end{array}
\]

2. Calculate \(f_i x_i\) for each class
\[
\begin{array}{c|c|c|c}
\text{Class} & f_i & x_i & f_i x_i\\ \hline
0.00-0.04 & 4 & 0.02 & 4\times0.02 = 0.08\\
0.04-0.08 & 9 & 0.06 & 9\times0.06 = 0.54\\
0.08-0.12 & 9 & 0.10 & 9\times0.10 = 0.90\\
0.12-0.16 & 2 & 0.14 & 2\times0.14 = 0.28\\
0.16-0.20 & 4 & 0.18 & 4\times0.18 = 0.72\\
0.20-0.24 & 2 & 0.22 & 2\times0.22 = 0.44\\
\end{array}
\]

3. Add the frequencies and the products
\[
\sum f_i = 4+9+9+2+4+2 = 30
\]
\[
\sum f_i x_i = 0.08+0.54+0.90+0.28+0.72+0.44 = 2.96
\]

4. Compute the mean
\[
\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2.96}{30}=0.09866\text{ ppm}\approx 0.10\text{ ppm (to two decimal places)}
\]

5. Answer: The mean concentration of SO₂ in the air is approximately 0.10 ppm.

1. List the class intervals, frequencies and find the mid‑points:
\[\begin{array}{c|c|c}
\text{Class interval (days)} & \text{Frequency (students)} & \text{Mid‑point } x_i \\ \hline
0-6 & 11 & \frac{0+6}{2}=3 \\
6-10 & 10 & \frac{6+10}{2}=8 \\
10-14 & 7 & \frac{10+14}{2}=12 \\
14-20 & 4 & \frac{14+20}{2}=17 \\
20-28 & 4 & \frac{20+28}{2}=24 \\
28-38 & 3 & \frac{28+38}{2}=33 \\
38-40 & 1 & \frac{38+40}{2}=39 \\
\end{array}\]

2. Compute the product of each mid‑point with its frequency (\(x_i f_i\)):
\[\begin{aligned}
3 \times 11 &= 33 \\
8 \times 10 &= 80 \\
12 \times 7 &= 84 \\
17 \times 4 &= 68 \\
24 \times 4 &= 96 \\
33 \times 3 &= 99 \\
39 \times 1 &= 39 \\
\end{aligned}\]

3. Add all the products:
\[\sum x_i f_i = 33+80+84+68+96+99+39 = 499\]

4. Add all the frequencies (total number of students):
\[\sum f_i = 11+10+7+4+4+3+1 = 40\]

5. Calculate the mean using the formula \(\bar{x}=\frac{\sum x_i f_i}{\sum f_i}\):
\[\bar{x}=\frac{499}{40}=12.475\]

6. Round off to two decimal places (as is customary in CBSE examinations):
\[\boxed{12.48 \text{ days (approximately)}}\]

Thus, the average number of days a student was absent during the term is about 12.48 days.

1. Identify the class intervals and their frequencies
- 45–55 : f = 3
- 55–65 : f = 10
- 65–75 : f = 11
- 75–85 : f = 8
- 85–95 : f = 3
Total number of cities, \(N = 3+10+11+8+3 = 35\).

2. Find the class‑mark (mid‑point) of each interval
\[\text{Class‑mark} = \frac{\text{lower limit}+\text{upper limit}}{2}\]
- 45–55 : \(x = \frac{45+55}{2}=50\)
- 55–65 : \(x = \frac{55+65}{2}=60\)
- 65–75 : \(x = \frac{65+75}{2}=70\)
- 75–85 : \(x = \frac{75+85}{2}=80\)
- 85–95 : \(x = \frac{85+95}{2}=90\)

3. Calculate \(f\times x\) for each class
\[
\begin{aligned}
50\times3 &= 150\\
60\times10 &= 600\\
70\times11 &= 770\\
80\times8 &= 640\\
90\times3 &= 270
\end{aligned}
\]
Sum of \(f\times x\) = \(150+600+770+640+270 = 2430\).

4. Compute the mean
\[
\bar{x} = \frac{\sum f x}{N} = \frac{2430}{35} \approx 69.4286\%\]
Rounded to two decimal places, the mean literacy rate is \(69.43\%\).

5. Answer: The mean literacy rate of the 35 cities is approximately 69.43 %.