EXERCISE 13.3
Surface Areas And Volumes • 7 Questions
Question 1
Hint available
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units) Number of consumers 65 - 85 4 85 - 105 5 105 - 125 13 125 - 145 20 145 - 165 14 165 - 185 8 185 - 205 4
Key Idea
For grouped data, the median is obtained using the formula \(L+\frac{\frac{N}{2}-c_f}{f_m}\times h\). The mean is calculated by taking class‑midpoints as representative values: \(\bar{x}=\frac{\sum f_i x_i}{N}\). The mode for grouped data uses \(L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h\), where \(f_1\) is the highest frequency, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively.
Step-by-Step Solution
1. Tabulate the data
| Class (units) | Frequency (f) | Mid‑point (x) |
|---------------|--------------|--------------|
| 65 – 85 | 4 | 75 |
| 85 – 105 | 5 | 95 |
| 105 – 125 | 13 | 115 |
| 125 – 145 | 20 | 135 |
| 145 – 165 | 14 | 155 |
| 165 – 185 | 8 | 175 |
| 185 – 205 | 4 | 195 |
Total \(N = 68\).
2. Median
- Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68.
- \(\frac{N}{2}=34\). The first cumulative frequency ≥ 34 is 42, so the median class is 125 – 145.
- Lower limit \(L = 125\), class width \(h = 20\), frequency of median class \(f_m = 20\), cumulative frequency before median class \(c_f = 22\).
- \[\text{Median}=L+\frac{\frac{N}{2}-c_f}{f_m}\times h =125+\frac{34-22}{20}\times20 =125+\frac{12}{20}\times20 =125+12 = \mathbf{137\;units}\]
3. Mean
- Compute \(\sum f_i x_i\):
\(75\times4=300\)
\(95\times5=475\)
\(115\times13=1495\)
\(135\times20=2700\)
\(155\times14=2170\)
\(175\times8=1400\)
\(195\times4=780\)
- Sum = \(300+475+1495+2700+2170+1400+780 = 9320\).
- \[\bar{x}=\frac{\sum f_i x_i}{N}=\frac{9320}{68}=\mathbf{137.06\;units\;(≈137)}\]
4. Mode
- Highest frequency = 20 (class 125 – 145). Hence modal class = 125 – 145.
- \(f_1 = 20\), \(f_0 = 13\) (preceding class), \(f_2 = 14\) (succeeding class).
- \[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h =125+\frac{20-13}{2\times20-13-14}\times20 =125+\frac{7}{13}\times20\]
- \(\frac{7}{13}=0.5385\). Hence \[\text{Mode}=125+0.5385\times20 =125+10.77 = \mathbf{135.77\;units\;(≈136)}\]
5. Comparison
- Mean \(\approx 137\), Median \(=137\), Mode \(\approx 136\).
- All three measures are very close, indicating that the distribution of electricity consumption is nearly symmetric about the central value.
- Hence, the data set is fairly balanced with no extreme skewness.
| Class (units) | Frequency (f) | Mid‑point (x) |
|---------------|--------------|--------------|
| 65 – 85 | 4 | 75 |
| 85 – 105 | 5 | 95 |
| 105 – 125 | 13 | 115 |
| 125 – 145 | 20 | 135 |
| 145 – 165 | 14 | 155 |
| 165 – 185 | 8 | 175 |
| 185 – 205 | 4 | 195 |
Total \(N = 68\).
2. Median
- Cumulative frequencies: 4, 9, 22, 42, 56, 64, 68.
- \(\frac{N}{2}=34\). The first cumulative frequency ≥ 34 is 42, so the median class is 125 – 145.
- Lower limit \(L = 125\), class width \(h = 20\), frequency of median class \(f_m = 20\), cumulative frequency before median class \(c_f = 22\).
- \[\text{Median}=L+\frac{\frac{N}{2}-c_f}{f_m}\times h =125+\frac{34-22}{20}\times20 =125+\frac{12}{20}\times20 =125+12 = \mathbf{137\;units}\]
3. Mean
- Compute \(\sum f_i x_i\):
\(75\times4=300\)
\(95\times5=475\)
\(115\times13=1495\)
\(135\times20=2700\)
\(155\times14=2170\)
\(175\times8=1400\)
\(195\times4=780\)
- Sum = \(300+475+1495+2700+2170+1400+780 = 9320\).
- \[\bar{x}=\frac{\sum f_i x_i}{N}=\frac{9320}{68}=\mathbf{137.06\;units\;(≈137)}\]
4. Mode
- Highest frequency = 20 (class 125 – 145). Hence modal class = 125 – 145.
- \(f_1 = 20\), \(f_0 = 13\) (preceding class), \(f_2 = 14\) (succeeding class).
- \[\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h =125+\frac{20-13}{2\times20-13-14}\times20 =125+\frac{7}{13}\times20\]
- \(\frac{7}{13}=0.5385\). Hence \[\text{Mode}=125+0.5385\times20 =125+10.77 = \mathbf{135.77\;units\;(≈136)}\]
5. Comparison
- Mean \(\approx 137\), Median \(=137\), Mode \(\approx 136\).
- All three measures are very close, indicating that the distribution of electricity consumption is nearly symmetric about the central value.
- Hence, the data set is fairly balanced with no extreme skewness.
Question 2
Hint available
If the median of the distribution given below is 28.5, find the values of x and y. Class interval Frequency 0 - 10 5 10 - 20 x 20 - 30 20 30 - 40 15 40 - 50 y 50 - 60 5 Total 60
Key Idea
For grouped data, the median is found using the formula \(M = L + \frac{\frac{N}{2} - C_f}{f}\,h\), where \(L\) is the lower limit of the median class, \(h\) is the class width, \(f\) is the frequency of the median class and \(C_f\) is the cumulative frequency before the median class. The total frequency gives a relation between the unknown frequencies.
Step-by-Step Solution
1. Total frequency
\[5 + x + 20 + 15 + y + 5 = 60 \]
\[45 + x + y = 60 \]
\[x + y = 15 \] (1)
2. Position of the median
\[N = 60 \Rightarrow \frac{N}{2}=30\]
The median is the 30th observation.
3. Cumulative frequencies
\[\begin{aligned}
\text{0-10}: &\; C_f = 5\\
\text{10-20}: &\; C_f = 5 + x\\
\text{20-30}: &\; C_f = 5 + x + 20 = 25 + x\\
\text{30-40}: &\; C_f = 40 + x\
\end{aligned}\]
Since the 30th observation must lie in the class where the cumulative frequency just exceeds 30, we need \(25 + x \ge 30\) i.e. \(x \ge 5\). Hence the median class is 20‑30.
4. Apply the median formula
\[L = 20,\; h = 10,\; f = 20,\; C_f = 5 + x\]
\[\begin{aligned}
28.5 &= 20 + \frac{30 - (5 + x)}{20}\times 10\\
28.5 - 20 &= \frac{25 - x}{20}\times 10\\
8.5 &= \frac{25 - x}{2}\\
25 - x &= 17\\
x &= 8\
\end{aligned}\]
5. Find y using (1)
\[y = 15 - x = 15 - 8 = 7\]
6. Verification
Cumulative frequency before the median class = 5 + 8 = 13 < 30,
Cumulative frequency after the median class = 13 + 20 = 33 ≥ 30, confirming that the median indeed lies in the 20‑30 class.
\[5 + x + 20 + 15 + y + 5 = 60 \]
\[45 + x + y = 60 \]
\[x + y = 15 \] (1)
2. Position of the median
\[N = 60 \Rightarrow \frac{N}{2}=30\]
The median is the 30th observation.
3. Cumulative frequencies
\[\begin{aligned}
\text{0-10}: &\; C_f = 5\\
\text{10-20}: &\; C_f = 5 + x\\
\text{20-30}: &\; C_f = 5 + x + 20 = 25 + x\\
\text{30-40}: &\; C_f = 40 + x\
\end{aligned}\]
Since the 30th observation must lie in the class where the cumulative frequency just exceeds 30, we need \(25 + x \ge 30\) i.e. \(x \ge 5\). Hence the median class is 20‑30.
4. Apply the median formula
\[L = 20,\; h = 10,\; f = 20,\; C_f = 5 + x\]
\[\begin{aligned}
28.5 &= 20 + \frac{30 - (5 + x)}{20}\times 10\\
28.5 - 20 &= \frac{25 - x}{20}\times 10\\
8.5 &= \frac{25 - x}{2}\\
25 - x &= 17\\
x &= 8\
\end{aligned}\]
5. Find y using (1)
\[y = 15 - x = 15 - 8 = 7\]
6. Verification
Cumulative frequency before the median class = 5 + 8 = 13 < 30,
Cumulative frequency after the median class = 13 + 20 = 33 ≥ 30, confirming that the median indeed lies in the 20‑30 class.
Question 3
Hint available
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. STATISTICS 199 Age (in years) Number of policy holders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100
Key Idea
For grouped data, the median is found using the formula \(\displaystyle \text{Median}=L+\frac{\frac{N}{2}-c_f}{f}\times h\), where \(L\) is the lower class boundary of the median class, \(c_f\) is the cumulative frequency before the median class, \(f\) is the frequency of the median class, \(h\) is the class width and \(N\) is the total number of observations.
Step-by-Step Solution
1. Total number of policy holders \(N = 100\).
2. Position of the median in the ordered data: \(\frac{N}{2}=\frac{100}{2}=50\) (the 50th observation).
3. Identify the median class using the cumulative frequencies:
- Cumulative frequency just before 35 years = 45 ("Below 35").
- Cumulative frequency just after 35 years = 78 ("Below 40").
Since 45 < 50 ≤ 78, the median lies in the class 35–40 years.
4. Extract required values:
- Lower class boundary \(L = 35\).
- Class width \(h = 40-35 = 5\) years.
- Cumulative frequency before the median class \(c_f = 45\).
- Frequency of the median class \(f = 78-45 = 33\).
5. Apply the median formula:
\[\text{Median}= L + \frac{\frac{N}{2} - c_f}{f}\times h \]
\[\text{Median}= 35 + \frac{50 - 45}{33}\times 5 \]
\[\text{Median}= 35 + \frac{5}{33}\times 5 \]
\[\text{Median}= 35 + \frac{25}{33} \]
\[\text{Median}= 35 + 0.7576 \approx 35.76\text{ years}\]
6. Result: The median age of the policy holders (aged 18 – < 60) is approximately 35.8 years.
*Note*: The lower age limit of 18 years does not affect the calculation because the median class (35–40) lies well within the given range.
2. Position of the median in the ordered data: \(\frac{N}{2}=\frac{100}{2}=50\) (the 50th observation).
3. Identify the median class using the cumulative frequencies:
- Cumulative frequency just before 35 years = 45 ("Below 35").
- Cumulative frequency just after 35 years = 78 ("Below 40").
Since 45 < 50 ≤ 78, the median lies in the class 35–40 years.
4. Extract required values:
- Lower class boundary \(L = 35\).
- Class width \(h = 40-35 = 5\) years.
- Cumulative frequency before the median class \(c_f = 45\).
- Frequency of the median class \(f = 78-45 = 33\).
5. Apply the median formula:
\[\text{Median}= L + \frac{\frac{N}{2} - c_f}{f}\times h \]
\[\text{Median}= 35 + \frac{50 - 45}{33}\times 5 \]
\[\text{Median}= 35 + \frac{5}{33}\times 5 \]
\[\text{Median}= 35 + \frac{25}{33} \]
\[\text{Median}= 35 + 0.7576 \approx 35.76\text{ years}\]
6. Result: The median age of the policy holders (aged 18 – < 60) is approximately 35.8 years.
*Note*: The lower age limit of 18 years does not affect the calculation because the median class (35–40) lies well within the given range.
Question 4
Hint available
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table : Length (in mm) Number of leaves 118 - 126 3 127 - 135 5 136 - 144 9 145 - 153 12 154 - 162 5 163 - 171 4 172 - 180 2 Find the median length of the leaves. (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.) 200
Key Idea
For grouped (continuous) data, the median is found using the formula \(\displaystyle \text{Median}=L+\frac{\frac{N}{2}-c_f}{f}\times h\), where \(L\) is the lower limit of the median class, \(c_f\) is the cumulative frequency before the median class, \(f\) is the frequency of the median class and \(h\) is the class width.
Step-by-Step Solution
1. Convert to continuous classes\
Original class \(118-126\) becomes \(117.5-126.5\), \(127-135\) becomes \(126.5-135.5\), and so on. The class width \(h\) is \(126.5-117.5 = 9\) mm.\
2. Tabulate frequencies and cumulative frequencies\
| Class (mm) | Frequency \(f\) | Cumulative \(c_f\) |
|------------|----------------|-------------------|
| 117.5‑126.5 | 3 | 3 |
| 126.5‑135.5 | 5 | 8 |
| 135.5‑144.5 | 9 | 17 |
| 144.5‑153.5 | 12 | 29 |
| 153.5‑162.5 | 5 | 34 |
| 162.5‑171.5 | 4 | 38 |
| 171.5‑180.5 | 2 | 40 |
Total \(N = 40\).\
3. Locate the median class\
Median position = \(\frac{N}{2}=20\) (or \(\frac{N+1}{2}=20.5\)). The cumulative frequency just before 20 is 17 (after the third class) and after the fourth class it becomes 29. Hence the median class is \(144.5‑153.5\).\
4. Apply the median formula\
\[\text{Median}=L+\frac{\frac{N}{2}-c_f}{f}\times h\]\
Here, \(L = 144.5\) mm, \(c_f = 17\), \(f = 12\), \(h = 9\) mm.\
\[\text{Median}=144.5+\frac{20-17}{12}\times 9\]
\[\text{Median}=144.5+\frac{3}{12}\times 9\]
\[\text{Median}=144.5+0.25\times 9\]
\[\text{Median}=144.5+2.25\]
\[\text{Median}=146.75\ \text{mm}\]
5. State the answer\
The median length of the leaves is \(146.75\) mm (≈ \(146.8\) mm to one decimal place).
Original class \(118-126\) becomes \(117.5-126.5\), \(127-135\) becomes \(126.5-135.5\), and so on. The class width \(h\) is \(126.5-117.5 = 9\) mm.\
2. Tabulate frequencies and cumulative frequencies\
| Class (mm) | Frequency \(f\) | Cumulative \(c_f\) |
|------------|----------------|-------------------|
| 117.5‑126.5 | 3 | 3 |
| 126.5‑135.5 | 5 | 8 |
| 135.5‑144.5 | 9 | 17 |
| 144.5‑153.5 | 12 | 29 |
| 153.5‑162.5 | 5 | 34 |
| 162.5‑171.5 | 4 | 38 |
| 171.5‑180.5 | 2 | 40 |
Total \(N = 40\).\
3. Locate the median class\
Median position = \(\frac{N}{2}=20\) (or \(\frac{N+1}{2}=20.5\)). The cumulative frequency just before 20 is 17 (after the third class) and after the fourth class it becomes 29. Hence the median class is \(144.5‑153.5\).\
4. Apply the median formula\
\[\text{Median}=L+\frac{\frac{N}{2}-c_f}{f}\times h\]\
Here, \(L = 144.5\) mm, \(c_f = 17\), \(f = 12\), \(h = 9\) mm.\
\[\text{Median}=144.5+\frac{20-17}{12}\times 9\]
\[\text{Median}=144.5+\frac{3}{12}\times 9\]
\[\text{Median}=144.5+0.25\times 9\]
\[\text{Median}=144.5+2.25\]
\[\text{Median}=146.75\ \text{mm}\]
5. State the answer\
The median length of the leaves is \(146.75\) mm (≈ \(146.8\) mm to one decimal place).
Question 5
Hint available
The following table gives the distribution of the life time of 400 neon lamps : Life time (in hours) Number of lamps 1500 - 2000 14 2000 - 2500 56 2500 - 3000 60 3000 - 3500 86 3500 - 4000 74 4000 - 4500 62 4500 - 5000 48 Find the median life time of a lamp.
Key Idea
For grouped data, the median is found using the formula \(\text{Median}=L+\left(\frac{\frac{N}{2}-c_f}{f}\right)h\), where \(L\) is the lower class boundary of the median class, \(c_f\) is the cumulative frequency before the median class, \(f\) is the frequency of the median class, \(h\) is the class width, and \(N\) is the total number of observations.
Step-by-Step Solution
1. Total number of lamps: \(N = 400\).
2. Find \(\frac{N}{2}\): \(\frac{N}{2}=200\).
3. Prepare cumulative frequencies:
- 1500‑2000: \(f=14\), \(c_f=0\), cumulative = 14
- 2000‑2500: \(f=56\), cumulative = 70
- 2500‑3000: \(f=60\), cumulative = 130
- 3000‑3500: \(f=86\), cumulative = 216
- 3500‑4000: \(f=74\), cumulative = 290
- 4000‑4500: \(f=62\), cumulative = 352
- 4500‑5000: \(f=48\), cumulative = 400
4. Locate the median class: The cumulative frequency just exceeds \(200\) in the class 3000‑3500 (cumulative = 216). Hence, median class = 3000‑3500.
5. Identify required quantities:
- Lower class boundary \(L = 3000\) (assuming continuous class limits).
- Class width \(h = 3500-3000 = 500\).
- Cumulative frequency before median class \(c_f = 130\).
- Frequency of median class \(f = 86\).
6. Apply the median formula:
\[
\text{Median}=L+\left(\frac{\frac{N}{2}-c_f}{f}\right)h
=3000+\left(\frac{200-130}{86}\right)\times 500
\]
\[
=3000+\left(\frac{70}{86}\right)\times 500
=3000+0.813953\times 500
\]
\[
=3000+406.98\approx 3407\text{ hours}
\]
7. Result: The median life time of a lamp is approximately 3407 hours (rounded to the nearest hour).
2. Find \(\frac{N}{2}\): \(\frac{N}{2}=200\).
3. Prepare cumulative frequencies:
- 1500‑2000: \(f=14\), \(c_f=0\), cumulative = 14
- 2000‑2500: \(f=56\), cumulative = 70
- 2500‑3000: \(f=60\), cumulative = 130
- 3000‑3500: \(f=86\), cumulative = 216
- 3500‑4000: \(f=74\), cumulative = 290
- 4000‑4500: \(f=62\), cumulative = 352
- 4500‑5000: \(f=48\), cumulative = 400
4. Locate the median class: The cumulative frequency just exceeds \(200\) in the class 3000‑3500 (cumulative = 216). Hence, median class = 3000‑3500.
5. Identify required quantities:
- Lower class boundary \(L = 3000\) (assuming continuous class limits).
- Class width \(h = 3500-3000 = 500\).
- Cumulative frequency before median class \(c_f = 130\).
- Frequency of median class \(f = 86\).
6. Apply the median formula:
\[
\text{Median}=L+\left(\frac{\frac{N}{2}-c_f}{f}\right)h
=3000+\left(\frac{200-130}{86}\right)\times 500
\]
\[
=3000+\left(\frac{70}{86}\right)\times 500
=3000+0.813953\times 500
\]
\[
=3000+406.98\approx 3407\text{ hours}
\]
7. Result: The median life time of a lamp is approximately 3407 hours (rounded to the nearest hour).
Question 6
Hint available
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters 1 - 4 4 - 7 7 - 10 10 - 13 13 - 16 16 - 19 Number of surnames 6 30 40 16 4 4 Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Key Idea
For grouped data, median, mean and mode are obtained using the formulas:
- Median: \(L + \frac{\frac{N}{2} - C_f}{f}\,h\)
- Mean: \(\bar{x}=\frac{\sum f_i x_i}{N}\) where \(x_i\) are class mid‑points.
- Mode (grouped): \(L + \frac{f_1-f_0}{(2f_1-f_0-f_2)}\,h\) where \(f_1\) is the frequency of the modal class, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively.
- Median: \(L + \frac{\frac{N}{2} - C_f}{f}\,h\)
- Mean: \(\bar{x}=\frac{\sum f_i x_i}{N}\) where \(x_i\) are class mid‑points.
- Mode (grouped): \(L + \frac{f_1-f_0}{(2f_1-f_0-f_2)}\,h\) where \(f_1\) is the frequency of the modal class, \(f_0\) and \(f_2\) are the frequencies of the preceding and succeeding classes respectively.
Step-by-Step Solution
1. Tabulate the data
| Class (letters) | Frequency (f) | Mid‑point (x) |
|-----------------|--------------|--------------|
| 1 – 4 | 6 | 2.5 |
| 4 – 7 | 30 | 5.5 |
| 7 – 10 | 40 | 8.5 |
| 10 – 13 | 16 | 11.5 |
| 13 – 16 | 4 | 14.5 |
| 16 – 19 | 4 | 17.5 |
Total \(N = 100\).
2. Median
- Cumulative frequencies: 6, 36, 76, 92, 96, 100.
- Position of median = \(\frac{N}{2}=50\).
- The 50th observation lies in the class 7 – 10 (cumulative just exceeds 50).
- \(L = 7\) (lower limit of median class), \(h = 3\) (class width), \(f = 40\) (frequency of median class), \(C_f = 36\) (cumulative frequency before median class).
- \[\text{Median}= L+\frac{\frac{N}{2}-C_f}{f}\,h = 7+\frac{50-36}{40}\times3 = 7+\frac{14}{40}\times3 = 7+1.05 = 8.05\]
Hence, the median number of letters \(\approx 8.05\).
3. Mean
- Compute \(\sum f_i x_i\):
\(6\times2.5 = 15\)
\(30\times5.5 = 165\)
\(40\times8.5 = 340\)
\(16\times11.5 = 184\)
\(4\times14.5 = 58\)
\(4\times17.5 = 70\)
\(\sum f_i x_i = 15+165+340+184+58+70 = 832\).
- Mean \(\bar{x}=\frac{\sum f_i x_i}{N}=\frac{832}{100}=8.32\).
Hence, the average (mean) number of letters is \(8.32\).
4. Mode (grouped data)
- The modal class is the one with highest frequency: 7 – 10 (frequency \(f_1=40\)).
- Frequencies of adjacent classes: \(f_0 = 30\) (previous class), \(f_2 = 16\) (next class).
- Using the modal formula:
\[\text{Mode}= L+\frac{f_1-f_0}{(2f_1-f_0-f_2)}\,h\]
\[= 7+\frac{40-30}{(2\times40-30-16)}\times3 = 7+\frac{10}{34}\times3\]
\[= 7+0.2941\times3 = 7+0.8823 \approx 7.88\]
Hence, the modal size of the surnames is about \(7.88\) letters.
Answers
- Median \(\approx 8.05\) letters
- Mean \(= 8.32\) letters
- Mode \(\approx 7.88\) letters
| Class (letters) | Frequency (f) | Mid‑point (x) |
|-----------------|--------------|--------------|
| 1 – 4 | 6 | 2.5 |
| 4 – 7 | 30 | 5.5 |
| 7 – 10 | 40 | 8.5 |
| 10 – 13 | 16 | 11.5 |
| 13 – 16 | 4 | 14.5 |
| 16 – 19 | 4 | 17.5 |
Total \(N = 100\).
2. Median
- Cumulative frequencies: 6, 36, 76, 92, 96, 100.
- Position of median = \(\frac{N}{2}=50\).
- The 50th observation lies in the class 7 – 10 (cumulative just exceeds 50).
- \(L = 7\) (lower limit of median class), \(h = 3\) (class width), \(f = 40\) (frequency of median class), \(C_f = 36\) (cumulative frequency before median class).
- \[\text{Median}= L+\frac{\frac{N}{2}-C_f}{f}\,h = 7+\frac{50-36}{40}\times3 = 7+\frac{14}{40}\times3 = 7+1.05 = 8.05\]
Hence, the median number of letters \(\approx 8.05\).
3. Mean
- Compute \(\sum f_i x_i\):
\(6\times2.5 = 15\)
\(30\times5.5 = 165\)
\(40\times8.5 = 340\)
\(16\times11.5 = 184\)
\(4\times14.5 = 58\)
\(4\times17.5 = 70\)
\(\sum f_i x_i = 15+165+340+184+58+70 = 832\).
- Mean \(\bar{x}=\frac{\sum f_i x_i}{N}=\frac{832}{100}=8.32\).
Hence, the average (mean) number of letters is \(8.32\).
4. Mode (grouped data)
- The modal class is the one with highest frequency: 7 – 10 (frequency \(f_1=40\)).
- Frequencies of adjacent classes: \(f_0 = 30\) (previous class), \(f_2 = 16\) (next class).
- Using the modal formula:
\[\text{Mode}= L+\frac{f_1-f_0}{(2f_1-f_0-f_2)}\,h\]
\[= 7+\frac{40-30}{(2\times40-30-16)}\times3 = 7+\frac{10}{34}\times3\]
\[= 7+0.2941\times3 = 7+0.8823 \approx 7.88\]
Hence, the modal size of the surnames is about \(7.88\) letters.
Answers
- Median \(\approx 8.05\) letters
- Mean \(= 8.32\) letters
- Mode \(\approx 7.88\) letters
Question 7
Hint available
The distribution below gives the weights of 30 students of a class. Find the median weight of the students. Weight (in kg) 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 Number of students 2 3 8 6 6 3 2
Key Idea
For grouped data, the median is found using the formula \(\displaystyle \text{Median}=L+\left(\frac{\frac{N}{2}-c_f}{f}\right)h\), where \(L\) is the lower class boundary of the median class, \(N\) is the total frequency, \(c_f\) is the cumulative frequency of the class preceding the median class, \(f\) is the frequency of the median class, and \(h\) is the class width.
Step-by-Step Solution
1. Total number of students: \(N = 30\).
2. Find \(\frac{N}{2}\): \(\frac{N}{2}=\frac{30}{2}=15\).
3. Prepare cumulative frequency table:
| Weight (kg) | Frequency (f) | Cumulative Frequency (c_f) |
|-------------|---------------|----------------------------|
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 2 + 3 = 5 |
| 50 – 55 | 8 | 5 + 8 = 13 |
| 55 – 60 | 6 | 13 + 6 = 19 |
| 60 – 65 | 6 | 19 + 6 = 25 |
| 65 – 70 | 3 | 25 + 3 = 28 |
| 70 – 75 | 2 | 28 + 2 = 30 |
4. Locate the median class: The median class is the first class whose cumulative frequency \(\ge 15\). Here, \(c_f = 19\) for the class 55 – 60 kg, so this is the median class.
5. Identify required values:
- Lower class boundary \(L = 55\) (since the class is 55–60 kg, the lower boundary is 55).
- Class width \(h = 60 - 55 = 5\) kg.
- Frequency of median class \(f = 6\).
- Cumulative frequency of the preceding class \(c_f = 13\).
6. Apply the median formula:
\[\text{Median}= L + \left(\frac{\frac{N}{2} - c_f}{f}\right) h\]
Substituting the values:
\[\text{Median}= 55 + \left(\frac{15 - 13}{6}\right) \times 5\]
\[\text{Median}= 55 + \left(\frac{2}{6}\right) \times 5\]
\[\text{Median}= 55 + \frac{1}{3} \times 5\]
\[\text{Median}= 55 + \frac{5}{3}\]
\[\text{Median}= 55 + 1.666\ldots \approx 56.7\text{ kg}\]
7. Result: The median weight of the 30 students is approximately 56.7 kg.
2. Find \(\frac{N}{2}\): \(\frac{N}{2}=\frac{30}{2}=15\).
3. Prepare cumulative frequency table:
| Weight (kg) | Frequency (f) | Cumulative Frequency (c_f) |
|-------------|---------------|----------------------------|
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 2 + 3 = 5 |
| 50 – 55 | 8 | 5 + 8 = 13 |
| 55 – 60 | 6 | 13 + 6 = 19 |
| 60 – 65 | 6 | 19 + 6 = 25 |
| 65 – 70 | 3 | 25 + 3 = 28 |
| 70 – 75 | 2 | 28 + 2 = 30 |
4. Locate the median class: The median class is the first class whose cumulative frequency \(\ge 15\). Here, \(c_f = 19\) for the class 55 – 60 kg, so this is the median class.
5. Identify required values:
- Lower class boundary \(L = 55\) (since the class is 55–60 kg, the lower boundary is 55).
- Class width \(h = 60 - 55 = 5\) kg.
- Frequency of median class \(f = 6\).
- Cumulative frequency of the preceding class \(c_f = 13\).
6. Apply the median formula:
\[\text{Median}= L + \left(\frac{\frac{N}{2} - c_f}{f}\right) h\]
Substituting the values:
\[\text{Median}= 55 + \left(\frac{15 - 13}{6}\right) \times 5\]
\[\text{Median}= 55 + \left(\frac{2}{6}\right) \times 5\]
\[\text{Median}= 55 + \frac{1}{3} \times 5\]
\[\text{Median}= 55 + \frac{5}{3}\]
\[\text{Median}= 55 + 1.666\ldots \approx 56.7\text{ kg}\]
7. Result: The median weight of the 30 students is approximately 56.7 kg.