EXERCISE 13.2
Surface Areas And Volumes • 6 Questions
Question 1
Hint available
The following table shows the ages of the patients admitted in a hospital during a year: Age (in years) 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 Number of patients 6 11 21 23 14 5 Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Key Idea
For grouped data, the mode is obtained from the modal class using the formula \(\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\), where \(L\) is the lower limit of the modal class, \(f_1\) its frequency, \(f_0\) the frequency of the preceding class, \(f_2\) the frequency of the succeeding class and \(h\) the class width. The mean is calculated as \(\bar{x}=\frac{\sum f\,x}{\sum f}\), where \(x\) is the class midpoint.
Step-by-Step Solution
1. Identify the class intervals and frequencies
\[\begin{array}{c|c|c}
\text{Class (years)} & \text{Midpoint }(x) & \text{Frequency }(f)\\ \hline
5-15 & 10 & 6\\
15-25 & 20 & 11\\
25-35 & 30 & 21\\
35-45 & 40 & 23\\
45-55 & 50 & 14\\
55-65 & 60 & 5\\
\end{array}\]
2. Mode
- The highest frequency is \(f_1=23\) for the class \(35-45\); therefore this is the modal class.
- \(f_0\) (frequency of the preceding class) = 21.
- \(f_2\) (frequency of the succeeding class) = 14.
- Lower limit \(L\) of the modal class = 35.
- Class width \(h\) = 10 (all classes have equal width).
Apply the modal formula:
\[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
=35+\frac{23-21}{2\times23-21-14}\times10
=35+\frac{2}{46-35}\times10
=35+\frac{2}{11}\times10
=35+\frac{20}{11}\approx36.8\text{ years}\]
3. Mean
- Compute \(f\times x\) for each class:
\[\begin{aligned}
6\times10 &= 60\\
11\times20 &= 220\\
21\times30 &= 630\\
23\times40 &= 920\\
14\times50 &= 700\\
5\times60 &= 300\\
\end{aligned}\]
- Sum of \(f\times x\): \(\sum f x = 60+220+630+920+700+300 = 2830\).
- Total number of patients: \(\sum f = 6+11+21+23+14+5 = 80\).
- Mean:
\[\bar{x}=\frac{\sum f x}{\sum f}=\frac{2830}{80}=35.375\approx35.4\text{ years}\]
4. Comparison and Interpretation
- The mode (≈ 36.8 years) indicates that the most frequent age group is the interval 35‑45 years.
- The mean (≈ 35.4 years) gives the average age of all patients.
- Since the mode is slightly greater than the mean, the distribution is slightly negatively (left) skewed – there are a few younger patients pulling the average down, while the largest concentration of patients lies a little higher, around 36‑37 years.
- Both measures are close, showing that the data are fairly symmetric with a mild skewness.
\[\begin{array}{c|c|c}
\text{Class (years)} & \text{Midpoint }(x) & \text{Frequency }(f)\\ \hline
5-15 & 10 & 6\\
15-25 & 20 & 11\\
25-35 & 30 & 21\\
35-45 & 40 & 23\\
45-55 & 50 & 14\\
55-65 & 60 & 5\\
\end{array}\]
2. Mode
- The highest frequency is \(f_1=23\) for the class \(35-45\); therefore this is the modal class.
- \(f_0\) (frequency of the preceding class) = 21.
- \(f_2\) (frequency of the succeeding class) = 14.
- Lower limit \(L\) of the modal class = 35.
- Class width \(h\) = 10 (all classes have equal width).
Apply the modal formula:
\[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
=35+\frac{23-21}{2\times23-21-14}\times10
=35+\frac{2}{46-35}\times10
=35+\frac{2}{11}\times10
=35+\frac{20}{11}\approx36.8\text{ years}\]
3. Mean
- Compute \(f\times x\) for each class:
\[\begin{aligned}
6\times10 &= 60\\
11\times20 &= 220\\
21\times30 &= 630\\
23\times40 &= 920\\
14\times50 &= 700\\
5\times60 &= 300\\
\end{aligned}\]
- Sum of \(f\times x\): \(\sum f x = 60+220+630+920+700+300 = 2830\).
- Total number of patients: \(\sum f = 6+11+21+23+14+5 = 80\).
- Mean:
\[\bar{x}=\frac{\sum f x}{\sum f}=\frac{2830}{80}=35.375\approx35.4\text{ years}\]
4. Comparison and Interpretation
- The mode (≈ 36.8 years) indicates that the most frequent age group is the interval 35‑45 years.
- The mean (≈ 35.4 years) gives the average age of all patients.
- Since the mode is slightly greater than the mean, the distribution is slightly negatively (left) skewed – there are a few younger patients pulling the average down, while the largest concentration of patients lies a little higher, around 36‑37 years.
- Both measures are close, showing that the data are fairly symmetric with a mild skewness.
Question 2
Hint available
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Lifetimes (in hours) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 Frequency 10 35 52 61 38 29 Determine the modal lifetimes of the components.
Key Idea
The modal class is the class interval having the highest frequency. For grouped data, the mode (modal value) is estimated using the formula: $$L = L_0 + \frac{f_1 - f_0}{(2f_1 - f_0 - f_2)}\,h$$ where \(L_0\) is the lower limit of the modal class, \(f_1\) is its frequency, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively, and \(h\) is the class width.
Step-by-Step Solution
1. Identify the frequencies for each class:
- 0‑20 : 10
- 20‑40 : 35
- 40‑60 : 52
- 60‑80 : 61
- 80‑100 : 38
- 100‑120 : 29
2. Find the modal class – the class with the greatest frequency. The maximum frequency is 61, which belongs to the class 60‑80.
3. Write down the required quantities for the mode formula:
- Lower limit of modal class, \(L_0\) = 60
- Class width, \(h\) = 20 (since each class interval is of equal width)
- Frequency of modal class, \(f_1\) = 61
- Frequency of preceding class, \(f_0\) = 52 (class 40‑60)
- Frequency of succeeding class, \(f_2\) = 38 (class 80‑100)
4. Apply the mode formula for grouped data:
$$L = 60 + \frac{61 - 52}{2\times61 - 52 - 38}\times 20$$
$$L = 60 + \frac{9}{122 - 90}\times 20$$
$$L = 60 + \frac{9}{32}\times 20$$
$$L = 60 + 0.28125 \times 20$$
$$L = 60 + 5.625$$
$$L = 65.625\text{ hours}$$
5. State the modal lifetime (rounded to one decimal place):
$$\boxed{\text{Modal lifetime } \approx 65.6 \text{ hours}}$$
- 0‑20 : 10
- 20‑40 : 35
- 40‑60 : 52
- 60‑80 : 61
- 80‑100 : 38
- 100‑120 : 29
2. Find the modal class – the class with the greatest frequency. The maximum frequency is 61, which belongs to the class 60‑80.
3. Write down the required quantities for the mode formula:
- Lower limit of modal class, \(L_0\) = 60
- Class width, \(h\) = 20 (since each class interval is of equal width)
- Frequency of modal class, \(f_1\) = 61
- Frequency of preceding class, \(f_0\) = 52 (class 40‑60)
- Frequency of succeeding class, \(f_2\) = 38 (class 80‑100)
4. Apply the mode formula for grouped data:
$$L = 60 + \frac{61 - 52}{2\times61 - 52 - 38}\times 20$$
$$L = 60 + \frac{9}{122 - 90}\times 20$$
$$L = 60 + \frac{9}{32}\times 20$$
$$L = 60 + 0.28125 \times 20$$
$$L = 60 + 5.625$$
$$L = 65.625\text{ hours}$$
5. State the modal lifetime (rounded to one decimal place):
$$\boxed{\text{Modal lifetime } \approx 65.6 \text{ hours}}$$
Question 3
Hint available
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure : Expenditure (in `) Number of families 1000 - 1500 24 1500 - 2000 40 2000 - 2500 33 2500 - 3000 28 3000 - 3500 30 3500 - 4000 22 4000 - 4500 16 4500 - 5000 7 STATISTICS 187
Key Idea
For grouped data, the modal class is the class with the highest frequency. The mode is estimated using the formula \(\displaystyle \text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h\), where \(L\) is the lower limit of the modal class, \(f_1\) its frequency, \(f_0\) the frequency of the preceding class, \(f_2\) the frequency of the succeeding class and \(h\) the class width. The mean for grouped data is obtained by taking the class mid‑points as representative values, multiplying each by its frequency, summing these products and dividing by the total number of observations.
Step-by-Step Solution
1. Identify the modal class\
The frequencies are: 24, 40, 33, 28, 30, 22, 16, 7.\
The largest frequency is 40, so the modal class is \(1500\!\!\text{–}\!2000\).
2. Compute the mode (grouped data formula)\
\[\begin{aligned}
L &= 1500 \text{ (lower limit of modal class)}\\
f_1 &= 40 \text{ (frequency of modal class)}\\
f_0 &= 24 \text{ (frequency of previous class)}\\
f_2 &= 33 \text{ (frequency of next class)}\\
h &= 2000-1500 = 500 \text{ (class width)}\\[4pt]
\text{Mode} &= L + \frac{f_1-f_0}{2f_1-f_0-f_2}\,h \\
&= 1500 + \frac{40-24}{2\times40-24-33}\times500 \\
&= 1500 + \frac{16}{80-57}\times500 \\
&= 1500 + \frac{16}{23}\times500 \\
&= 1500 + 0.695652\times500 \\
&= 1500 + 347.83 \approx \mathbf{1848\text{ (Rs.)}}
\end{aligned}\]
Hence the modal monthly expenditure is approximately Rs. 1850.
3. Find the mean expenditure\
- Compute the class mid‑points (\(x_i\)):\
\[\begin{array}{c|c}
\text{Class} & \text{Mid‑point } x_i \\ \hline
1000-1500 & 1250 \\
1500-2000 & 1750 \\
2000-2500 & 2250 \\
2500-3000 & 2750 \\
3000-3500 & 3250 \\
3500-4000 & 3750 \\
4000-4500 & 4250 \\
4500-5000 & 4750 \\
\end{array}\]
- Multiply each mid‑point by its frequency (\(f_i\)) and sum:
\[\begin{aligned}
\sum f_i x_i &= 1250\times24 + 1750\times40 + 2250\times33 + 2750\times28 \\
&\quad + 3250\times30 + 3750\times22 + 4250\times16 + 4750\times7 \\
&= 30\,000 + 70\,000 + 74\,250 + 77\,000 \\
&\quad + 97\,500 + 82\,500 + 68\,000 + 33\,250 \\
&= 532\,500
\end{aligned}\]
- Total number of families \(N = 200\).
- Mean \(\bar{x}\):
\[\bar{x}=\frac{\sum f_i x_i}{N}=\frac{532\,500}{200}=\mathbf{2\,662.5\text{ (Rs.)}}\]
- Rounded to the nearest rupee, the mean monthly expenditure is Rs. 2663.
4. Answer\
- Modal monthly expenditure ≈ Rs. 1850.\
- Mean monthly expenditure ≈ Rs. 2663.
The frequencies are: 24, 40, 33, 28, 30, 22, 16, 7.\
The largest frequency is 40, so the modal class is \(1500\!\!\text{–}\!2000\).
2. Compute the mode (grouped data formula)\
\[\begin{aligned}
L &= 1500 \text{ (lower limit of modal class)}\\
f_1 &= 40 \text{ (frequency of modal class)}\\
f_0 &= 24 \text{ (frequency of previous class)}\\
f_2 &= 33 \text{ (frequency of next class)}\\
h &= 2000-1500 = 500 \text{ (class width)}\\[4pt]
\text{Mode} &= L + \frac{f_1-f_0}{2f_1-f_0-f_2}\,h \\
&= 1500 + \frac{40-24}{2\times40-24-33}\times500 \\
&= 1500 + \frac{16}{80-57}\times500 \\
&= 1500 + \frac{16}{23}\times500 \\
&= 1500 + 0.695652\times500 \\
&= 1500 + 347.83 \approx \mathbf{1848\text{ (Rs.)}}
\end{aligned}\]
Hence the modal monthly expenditure is approximately Rs. 1850.
3. Find the mean expenditure\
- Compute the class mid‑points (\(x_i\)):\
\[\begin{array}{c|c}
\text{Class} & \text{Mid‑point } x_i \\ \hline
1000-1500 & 1250 \\
1500-2000 & 1750 \\
2000-2500 & 2250 \\
2500-3000 & 2750 \\
3000-3500 & 3250 \\
3500-4000 & 3750 \\
4000-4500 & 4250 \\
4500-5000 & 4750 \\
\end{array}\]
- Multiply each mid‑point by its frequency (\(f_i\)) and sum:
\[\begin{aligned}
\sum f_i x_i &= 1250\times24 + 1750\times40 + 2250\times33 + 2750\times28 \\
&\quad + 3250\times30 + 3750\times22 + 4250\times16 + 4750\times7 \\
&= 30\,000 + 70\,000 + 74\,250 + 77\,000 \\
&\quad + 97\,500 + 82\,500 + 68\,000 + 33\,250 \\
&= 532\,500
\end{aligned}\]
- Total number of families \(N = 200\).
- Mean \(\bar{x}\):
\[\bar{x}=\frac{\sum f_i x_i}{N}=\frac{532\,500}{200}=\mathbf{2\,662.5\text{ (Rs.)}}\]
- Rounded to the nearest rupee, the mean monthly expenditure is Rs. 2663.
4. Answer\
- Modal monthly expenditure ≈ Rs. 1850.\
- Mean monthly expenditure ≈ Rs. 2663.
Question 4
Hint available
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher Number of states / U.T. 15 - 20 3 20 - 25 8 25 - 30 9 30 - 35 10 35 - 40 3 40 - 45 0 45 - 50 0 50 - 55 2
Key Idea
For grouped data, the mean is obtained using class‑midpoints (x̄ = Σf·x / Σf). The mode is found from the modal class using the formula: \(\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h\), where \(L\) is the lower class boundary of the modal class, \(f_1\) its frequency, \(f_0\) the preceding class frequency, \(f_2\) the succeeding class frequency and \(h\) the class width.
Step-by-Step Solution
1. Tabulate the data
| Class (students/teacher) | Frequency (f) | Mid‑point (x) |
|--------------------------|--------------|--------------|
| 15 – 20 | 3 | 17.5 |
| 20 – 25 | 8 | 22.5 |
| 25 – 30 | 9 | 27.5 |
| 30 – 35 | 10 | 32.5 |
| 35 – 40 | 3 | 37.5 |
| 40 – 45 | 0 | 42.5 |
| 45 – 50 | 0 | 47.5 |
| 50 – 55 | 2 | 52.5 |
2. Total frequency \(N = 3+8+9+10+3+0+0+2 = 35\).
3. Mean (grouped data)
Compute \(\sum f x\):
\(3\times17.5 = 52.5\)
\(8\times22.5 = 180\)
\(9\times27.5 = 247.5\)
\(10\times32.5 = 325\)
\(3\times37.5 = 112.5\)
\(2\times52.5 = 105\)
\(\sum f x = 52.5+180+247.5+325+112.5+105 = 1,022.5\)
\[\bar{x}=\frac{\sum f x}{N}=\frac{1,022.5}{35}=29.214\approx 29.2\]
Hence the mean teacher‑student ratio ≈ 29.2 students per teacher.
4. Mode (grouped data)
- Identify the modal class (largest frequency): 30 – 35 (\(f_1=10\)).
- Preceding class frequency \(f_0 = 9\) (25 – 30).
- Succeeding class frequency \(f_2 = 3\) (35 – 40).
- Class width \(h = 5\).
- Lower class boundary \(L = 30\).
- Apply the formula:
\[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h
=30+\frac{10-9}{2\times10-9-3}\times5
=30+\frac{1}{8}\times5
=30+0.625=30.625\]
Hence the mode ≈ 30.6 students per teacher.
5. Interpretation
- *Mode* (≈30.6) tells that the most frequently occurring teacher‑student ratio among the states lies in the 30‑35 interval; i.e., most states have about 31 students per teacher.
- *Mean* (≈29.2) gives the overall average ratio across all states. Since the mean is slightly lower than the mode, the distribution is mildly left‑skewed – a few states with lower ratios (e.g., the 15‑20 interval) pull the average down while the bulk of states cluster around the higher ratio indicated by the mode.
- Together, they indicate that while the typical state has around 31 students per teacher, on the whole India’s average is a little lower because of the presence of some states with comparatively better (lower) teacher‑student ratios.
| Class (students/teacher) | Frequency (f) | Mid‑point (x) |
|--------------------------|--------------|--------------|
| 15 – 20 | 3 | 17.5 |
| 20 – 25 | 8 | 22.5 |
| 25 – 30 | 9 | 27.5 |
| 30 – 35 | 10 | 32.5 |
| 35 – 40 | 3 | 37.5 |
| 40 – 45 | 0 | 42.5 |
| 45 – 50 | 0 | 47.5 |
| 50 – 55 | 2 | 52.5 |
2. Total frequency \(N = 3+8+9+10+3+0+0+2 = 35\).
3. Mean (grouped data)
Compute \(\sum f x\):
\(3\times17.5 = 52.5\)
\(8\times22.5 = 180\)
\(9\times27.5 = 247.5\)
\(10\times32.5 = 325\)
\(3\times37.5 = 112.5\)
\(2\times52.5 = 105\)
\(\sum f x = 52.5+180+247.5+325+112.5+105 = 1,022.5\)
\[\bar{x}=\frac{\sum f x}{N}=\frac{1,022.5}{35}=29.214\approx 29.2\]
Hence the mean teacher‑student ratio ≈ 29.2 students per teacher.
4. Mode (grouped data)
- Identify the modal class (largest frequency): 30 – 35 (\(f_1=10\)).
- Preceding class frequency \(f_0 = 9\) (25 – 30).
- Succeeding class frequency \(f_2 = 3\) (35 – 40).
- Class width \(h = 5\).
- Lower class boundary \(L = 30\).
- Apply the formula:
\[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h
=30+\frac{10-9}{2\times10-9-3}\times5
=30+\frac{1}{8}\times5
=30+0.625=30.625\]
Hence the mode ≈ 30.6 students per teacher.
5. Interpretation
- *Mode* (≈30.6) tells that the most frequently occurring teacher‑student ratio among the states lies in the 30‑35 interval; i.e., most states have about 31 students per teacher.
- *Mean* (≈29.2) gives the overall average ratio across all states. Since the mean is slightly lower than the mode, the distribution is mildly left‑skewed – a few states with lower ratios (e.g., the 15‑20 interval) pull the average down while the bulk of states cluster around the higher ratio indicated by the mode.
- Together, they indicate that while the typical state has around 31 students per teacher, on the whole India’s average is a little lower because of the presence of some states with comparatively better (lower) teacher‑student ratios.
Question 5
Hint available
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored Number of batsmen 3000 - 4000 4 4000 - 5000 18 5000 - 6000 9 6000 - 7000 7 7000 - 8000 6 8000 - 9000 3 9000 - 10000 1 10000 - 11000 1 Find the mode of the data.
Key Idea
For grouped data, the mode is estimated using the formula: $$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$$ where \(L\) is the lower boundary of the modal class, \(f_1\) is the frequency of the modal class, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively, and \(h\) is the class width.
Step-by-Step Solution
1. Identify the modal class – the class with the highest frequency.\
The frequencies are: 4, 18, 9, 7, 6, 3, 1, 1. Hence the modal class is 4000 – 5000.
2. Write down the required quantities:\
- Lower class boundary of the modal class, \(L = 4000\).
- Class width, \(h = 5000-4000 = 1000\).
- Frequency of modal class, \(f_1 = 18\).
- Frequency of the preceding class, \(f_0 = 4\) (class 3000–4000).
- Frequency of the succeeding class, \(f_2 = 9\) (class 5000–6000).
3. Apply the mode formula for grouped data:\
$$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$$
Substituting the values:
$$\text{Mode}=4000+\frac{18-4}{2\times18-4-9}\times1000$$
$$\text{Mode}=4000+\frac{14}{36-13}\times1000$$
$$\text{Mode}=4000+\frac{14}{23}\times1000$$
$$\text{Mode}=4000+608.7$$
$$\text{Mode}\approx 4608.7$$
4. State the answer – Rounding to the nearest whole number (as runs are counted in whole numbers), the mode is approximately 4609 runs.
Thus, the most frequently occurring run‑score interval is centred around ≈ 4609 runs.
The frequencies are: 4, 18, 9, 7, 6, 3, 1, 1. Hence the modal class is 4000 – 5000.
2. Write down the required quantities:\
- Lower class boundary of the modal class, \(L = 4000\).
- Class width, \(h = 5000-4000 = 1000\).
- Frequency of modal class, \(f_1 = 18\).
- Frequency of the preceding class, \(f_0 = 4\) (class 3000–4000).
- Frequency of the succeeding class, \(f_2 = 9\) (class 5000–6000).
3. Apply the mode formula for grouped data:\
$$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$$
Substituting the values:
$$\text{Mode}=4000+\frac{18-4}{2\times18-4-9}\times1000$$
$$\text{Mode}=4000+\frac{14}{36-13}\times1000$$
$$\text{Mode}=4000+\frac{14}{23}\times1000$$
$$\text{Mode}=4000+608.7$$
$$\text{Mode}\approx 4608.7$$
4. State the answer – Rounding to the nearest whole number (as runs are counted in whole numbers), the mode is approximately 4609 runs.
Thus, the most frequently occurring run‑score interval is centred around ≈ 4609 runs.
Question 6
Hint available
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data : Number of cars 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency 7 14 13 12 20 11 15 8 188
Key Idea
Mode of grouped (continuous) data is found using the formula: \(\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\), where \(l\) is the lower limit of the modal class, \(f_1\) is its frequency, \(f_0\) and \(f_2\) are frequencies of the preceding and succeeding classes respectively, and \(h\) is the class width.
Step-by-Step Solution
1. Identify the modal class
The class with the highest frequency is the modal class. From the table, the frequencies are: \(7, 14, 13, 12, 20, 11, 15, 8\). The maximum frequency is \(20\) which corresponds to the class \(40-50\). Hence, the modal class is \(40-50\).\
2. Write down the required quantities
- Lower limit of modal class, \(l = 40\)
- Class width, \(h = 10\) (since each class interval is of size 10)\
- Frequency of modal class, \(f_1 = 20\)\
- Frequency of preceding class, \(f_0 = 12\) (class \(30-40\))\
- Frequency of succeeding class, \(f_2 = 11\) (class \(50-60\))\
3. Apply the mode formula
\[\text{Mode}= l + \frac{f_1-f_0}{2f_1-f_0-f_2}\times h \]\
Substitute the values:
\[\text{Mode}= 40 + \frac{20-12}{2\times20-12-11}\times 10 \]\
\[\text{Mode}= 40 + \frac{8}{40-23}\times 10 \]\
\[\text{Mode}= 40 + \frac{8}{17}\times 10 \]\
\[\text{Mode}= 40 + \frac{80}{17} \]\
\[\text{Mode}= 40 + 4.7059 \approx 44.7 \]\
4. State the answer
The mode of the data (to one decimal place) is approximately 44.7 cars.
The class with the highest frequency is the modal class. From the table, the frequencies are: \(7, 14, 13, 12, 20, 11, 15, 8\). The maximum frequency is \(20\) which corresponds to the class \(40-50\). Hence, the modal class is \(40-50\).\
2. Write down the required quantities
- Lower limit of modal class, \(l = 40\)
- Class width, \(h = 10\) (since each class interval is of size 10)\
- Frequency of modal class, \(f_1 = 20\)\
- Frequency of preceding class, \(f_0 = 12\) (class \(30-40\))\
- Frequency of succeeding class, \(f_2 = 11\) (class \(50-60\))\
3. Apply the mode formula
\[\text{Mode}= l + \frac{f_1-f_0}{2f_1-f_0-f_2}\times h \]\
Substitute the values:
\[\text{Mode}= 40 + \frac{20-12}{2\times20-12-11}\times 10 \]\
\[\text{Mode}= 40 + \frac{8}{40-23}\times 10 \]\
\[\text{Mode}= 40 + \frac{8}{17}\times 10 \]\
\[\text{Mode}= 40 + \frac{80}{17} \]\
\[\text{Mode}= 40 + 4.7059 \approx 44.7 \]\
4. State the answer
The mode of the data (to one decimal place) is approximately 44.7 cars.