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Mode of Grouped Data
| Age (in years) | 5 - 15 | 15 - 25 | 25 - 35 | 35 - 45 | 45 - 55 | 55 - 65 |
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Answer:
Mode.
Maximum frequency = 23, so the modal class is 35 - 45.
\( l = 35,\ h = 10,\ f_1 = 23,\ f_0 = 21,\ f_2 = 14 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 35 + \frac{23 - 21}{2(23) - 21 - 14} \times 10 = 36.82 \).
Mean.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
| Class | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 5 - 15 | 6 | 10 | 60 |
| 15 - 25 | 11 | 20 | 220 |
| 25 - 35 | 21 | 30 | 630 |
| 35 - 45 | 23 | 40 | 920 |
| 45 - 55 | 14 | 50 | 700 |
| 55 - 65 | 5 | 60 | 300 |
| Total | \( \sum f_i = 80 \) | \( \sum f_ix_i = 2830 \) |
Mean \( = \frac{2830}{80} = 35.375 \approx 35.38 \) years.
Interpretation. The maximum number of patients admitted are of age about 36.8 years (the mode), while on an average the age of a patient admitted to the hospital is about 35.4 years (the mean).
In simple words: Find the modal class for the mode and the class marks for the mean, then compare the two values.
Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.
| Lifetime (in hours) | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Answer:
Maximum frequency = 61, so the modal class is 60 - 80.
\( l = 60,\ h = 20,\ f_1 = 61,\ f_0 = 52,\ f_2 = 38 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 60 + \frac{61 - 52}{2(61) - 52 - 38} \times 20 = 65.625 \).
Therefore the modal lifetime of the components is 65.625 hours.
In simple words: The class with the highest frequency (60 - 80) gives the mode.
Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.
| Expenditure (in Rs.) | 1000 - 1500 | 1500 - 2000 | 2000 - 2500 | 2500 - 3000 | 3000 - 3500 | 3500 - 4000 | 4000 - 4500 | 4500 - 5000 |
| Number of families | 24 | 40 | 33 | 28 | 30 | 22 | 16 | 7 |
Answer:
Mode.
Maximum frequency = 40, so the modal class is 1500 - 2000.
\( l = 1500,\ h = 500,\ f_1 = 40,\ f_0 = 24,\ f_2 = 33 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 1500 + \frac{40 - 24}{2(40) - 24 - 33} \times 500 = 1847.83 \).
Therefore the modal monthly expenditure is Rs. 1847.83.
Mean.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 2750 \), class size \( h = 500 \) and \( u_i = \frac{x_i - a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( u_i = \frac{x_i - a}{h} \) | \( f_iu_i \) |
|---|---|---|---|---|
| 1000 - 1500 | 24 | 1250 | -3 | -72 |
| 1500 - 2000 | 40 | 1750 | -2 | -80 |
| 2000 - 2500 | 33 | 2250 | -1 | -33 |
| 2500 - 3000 | 28 | 2750 | 0 | 0 |
| 3000 - 3500 | 30 | 3250 | 1 | 30 |
| 3500 - 4000 | 22 | 3750 | 2 | 44 |
| 4000 - 4500 | 16 | 4250 | 3 | 48 |
| 4500 - 5000 | 7 | 4750 | 4 | 28 |
| Total | \( \sum f_i = 200 \) | \( \sum f_iu_i = -35 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 2750 + 500 \times \frac{-35}{200} = 2662.5 \).
Therefore the mean monthly expenditure is Rs. 2662.50.
In simple words: Use the modal class for the mode and the step-deviation method for the mean.
Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.
| Number of students per teacher | 15 - 20 | 20 - 25 | 25 - 30 | 30 - 35 | 35 - 40 | 40 - 45 | 45 - 50 | 50 - 55 |
| Number of states / U.T. | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 |
Answer:
Mode.
Maximum frequency = 10, so the modal class is 30 - 35.
\( l = 30,\ h = 5,\ f_1 = 10,\ f_0 = 9,\ f_2 = 3 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 30 + \frac{10 - 9}{2(10) - 9 - 3} \times 5 = 30.625 \).
Mean.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 32.5 \), class size \( h = 5 \) and \( u_i = \frac{x_i - a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( u_i = \frac{x_i - a}{h} \) | \( f_iu_i \) |
|---|---|---|---|---|
| 15 - 20 | 3 | 17.5 | -3 | -9 |
| 20 - 25 | 8 | 22.5 | -2 | -16 |
| 25 - 30 | 9 | 27.5 | -1 | -9 |
| 30 - 35 | 10 | 32.5 | 0 | 0 |
| 35 - 40 | 3 | 37.5 | 1 | 3 |
| 40 - 45 | 0 | 42.5 | 2 | 0 |
| 45 - 50 | 0 | 47.5 | 3 | 0 |
| 50 - 55 | 2 | 52.5 | 4 | 8 |
| Total | \( \sum f_i = 35 \) | \( \sum f_iu_i = -23 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 32.5 + 5 \times \frac{-23}{35} = 29.21 \).
Interpretation. Most states/U.T. have a student-teacher ratio of about 30.6, while on average the ratio is about 29.2.
In simple words: Mode tells the most common ratio; mean tells the average ratio.
Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.
| Runs scored | 3000 - 4000 | 4000 - 5000 | 5000 - 6000 | 6000 - 7000 | 7000 - 8000 | 8000 - 9000 | 9000 - 10000 | 10000 - 11000 |
| Number of batsmen | 4 | 18 | 9 | 7 | 6 | 3 | 1 | 1 |
Answer:
Maximum frequency = 18, so the modal class is 4000 - 5000.
\( l = 4000,\ h = 1000,\ f_1 = 18,\ f_0 = 4,\ f_2 = 9 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 4000 + \frac{18 - 4}{2(18) - 4 - 9} \times 1000 = 4608.7 \).
Therefore the mode of the given data is about 4608.7 runs.
In simple words: Highest frequency 18 gives the modal class 4000 - 5000.
Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.
| 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 |
Answer:
Maximum frequency = 20, so the modal class is 40 - 50.
\( l = 40,\ h = 10,\ f_1 = 20,\ f_0 = 12,\ f_2 = 11 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 40 + \frac{20 - 12}{2(20) - 12 - 11} \times 10 = 44.71 \).
Therefore the mode of the given data is about 44.7 cars.
In simple words: The class 40 - 50 has the highest frequency 20.
Teacher's Note:
1. Locate the modal class first: it is the class with the highest frequency.
2. Take care to use f0 for the class before and f2 for the class after the modal class.
Median of Grouped Data
| Monthly consumption (in units) | 65 - 85 | 85 - 105 | 105 - 125 | 125 - 145 | 145 - 165 | 165 - 185 | 185 - 205 |
| Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 | 4 |
Answer:
Median.
| Class | Frequency | Cumulative frequency |
|---|---|---|
| 65 - 85 | 4 | 4 |
| 85 - 105 | 5 | 9 |
| 105 - 125 | 13 | 22 |
| 125 - 145 | 20 | 42 |
| 145 - 165 | 14 | 56 |
| 165 - 185 | 8 | 64 |
| 185 - 205 | 4 | 68 |
\( N = 68 \), so \( \frac{N}{2} = 34 \). The cumulative frequency just greater than this is 42, so the median class is 125 - 145.
\( l = 125,\ c.f. = 22,\ f = 20,\ h = 20 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 125 + \frac{34 - 22}{20} \times 20 = 137 \).
Mode.
Maximum frequency = 20, so the modal class is 125 - 145.
\( l = 125,\ h = 20,\ f_1 = 20,\ f_0 = 13,\ f_2 = 14 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 125 + \frac{20 - 13}{2(20) - 13 - 14} \times 20 = 135.77 \).
Mean.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \). Take \( a = 135 \), class size \( h = 20 \) and \( u_i = \frac{x_i - a}{h} \).
| Class | \( f_i \) | \( x_i \) | \( u_i = \frac{x_i - a}{h} \) | \( f_iu_i \) |
|---|---|---|---|---|
| 65 - 85 | 4 | 75 | -3 | -12 |
| 85 - 105 | 5 | 95 | -2 | -10 |
| 105 - 125 | 13 | 115 | -1 | -13 |
| 125 - 145 | 20 | 135 | 0 | 0 |
| 145 - 165 | 14 | 155 | 1 | 14 |
| 165 - 185 | 8 | 175 | 2 | 16 |
| 185 - 205 | 4 | 195 | 3 | 12 |
| Total | \( \sum f_i = 68 \) | \( \sum f_iu_i = 7 \) |
Mean \( \bar{x} = a + h \times \frac{\sum f_iu_i}{\sum f_i} = 135 + 20 \times \frac{7}{68} = 137.06 \).
The median, mean and mode (137, 137.06 and 135.77) are approximately equal in this distribution.
In simple words: Find the three measures separately; here they come out close to each other.
Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | Total |
| Frequency | 5 | x | 20 | 15 | y | 5 | 60 |
Answer:
| Class interval | Frequency | Cumulative frequency |
|---|---|---|
| 0 - 10 | 5 | 5 |
| 10 - 20 | x | \( 5 + x \) |
| 20 - 30 | 20 | \( 25 + x \) |
| 30 - 40 | 15 | \( 40 + x \) |
| 40 - 50 | y | \( 40 + x + y \) |
| 50 - 60 | 5 | \( 45 + x + y \) |
\( N = 60 \), so \( \frac{N}{2} = 30 \). The median is 28.5, so the median class is 20 - 30 with \( l = 20,\ c.f. = 5 + x,\ f = 20,\ h = 10 \).
\( 28.5 = 20 + \frac{30 - (5 + x)}{20} \times 10 \Rightarrow 8.5 = \frac{25 - x}{2} \Rightarrow x = 8 \).
Also \( 45 + x + y = 60 \Rightarrow y = 60 - 45 - 8 = 7 \). So x = 8 and y = 7.
In simple words: Use the median formula to find x, then the total frequency to find y.
Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.
| Age (in years) | Below 20 | Below 25 | Below 30 | Below 35 | Below 40 | Below 45 | Below 50 | Below 55 | Below 60 |
| Number of policy holders | 2 | 6 | 24 | 45 | 78 | 89 | 92 | 98 | 100 |
Answer:
The data is a cumulative frequency distribution. Since policies are given only from age 18, the classes are 18 - 20, 20 - 25, ..., 55 - 60, and each frequency is the difference of consecutive cumulative frequencies.
| Class | Frequency | Cumulative frequency |
|---|---|---|
| 18 - 20 | 2 | 2 |
| 20 - 25 | 4 | 6 |
| 25 - 30 | 18 | 24 |
| 30 - 35 | 21 | 45 |
| 35 - 40 | 33 | 78 |
| 40 - 45 | 11 | 89 |
| 45 - 50 | 3 | 92 |
| 50 - 55 | 6 | 98 |
| 55 - 60 | 2 | 100 |
\( N = 100 \), so \( \frac{N}{2} = 50 \). The cumulative frequency just greater than this is 78, so the median class is 35 - 40.
\( l = 35,\ c.f. = 45,\ f = 33,\ h = 5 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 35 + \frac{50 - 45}{33} \times 5 = 35.76 \).
Therefore the median age is about 35.76 years.
In simple words: Convert the 'below' data to class frequencies and then apply the median formula.
Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.
| Length (in mm) | 118-126 | 127-135 | 136-144 | 145-153 | 154-162 | 163-171 | 172-180 |
| Number of leaves | 3 | 5 | 9 | 12 | 5 | 4 | 2 |
Answer:
The classes are not continuous (a gap of 1 between classes). Subtract 0.5 from each lower limit and add 0.5 to each upper limit.
| Class | Frequency | Cumulative frequency |
|---|---|---|
| 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 |
\( N = 40 \), so \( \frac{N}{2} = 20 \). The cumulative frequency just greater than this is 29, so the median class is 144.5 - 153.5.
\( l = 144.5,\ c.f. = 17,\ f = 12,\ h = 9 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 144.5 + \frac{20 - 17}{12} \times 9 = 146.75 \).
Therefore the median length of the leaves is 146.75 mm.
In simple words: Make the classes continuous before using the median formula.
Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.
| Lifetime (in hours) | 1500 - 2000 | 2000 - 2500 | 2500 - 3000 | 3000 - 3500 | 3500 - 4000 | 4000 - 4500 | 4500 - 5000 |
| Number of lamps | 14 | 56 | 60 | 86 | 74 | 62 | 48 |
Answer:
| Class | Frequency | Cumulative frequency |
|---|---|---|
| 1500 - 2000 | 14 | 14 |
| 2000 - 2500 | 56 | 70 |
| 2500 - 3000 | 60 | 130 |
| 3000 - 3500 | 86 | 216 |
| 3500 - 4000 | 74 | 290 |
| 4000 - 4500 | 62 | 352 |
| 4500 - 5000 | 48 | 400 |
\( N = 400 \), so \( \frac{N}{2} = 200 \). The cumulative frequency just greater than this is 216, so the median class is 3000 - 3500.
\( l = 3000,\ c.f. = 130,\ f = 86,\ h = 500 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 3000 + \frac{200 - 130}{86} \times 500 = 3406.98 \).
Therefore the median lifetime of a lamp is about 3406.98 hours.
In simple words: Find N/2 = 200, locate the median class 3000 - 3500 and apply the formula.
Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.
| Number of letters | 1 - 4 | 4 - 7 | 7 - 10 | 10 - 13 | 13 - 16 | 16 - 19 |
| Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |
Answer:
Median.
| Class | Frequency | Cumulative frequency |
|---|---|---|
| 1 - 4 | 6 | 6 |
| 4 - 7 | 30 | 36 |
| 7 - 10 | 40 | 76 |
| 10 - 13 | 16 | 92 |
| 13 - 16 | 4 | 96 |
| 16 - 19 | 4 | 100 |
\( N = 100 \), so \( \frac{N}{2} = 50 \). The cumulative frequency just greater than this is 76, so the median class is 7 - 10.
\( l = 7,\ c.f. = 36,\ f = 40,\ h = 3 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 7 + \frac{50 - 36}{40} \times 3 = 8.05 \).
Mode.
Maximum frequency = 40, so the modal class is 7 - 10.
\( l = 7,\ h = 3,\ f_1 = 40,\ f_0 = 30,\ f_2 = 16 \).
Mode \( = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 7 + \frac{40 - 30}{2(40) - 30 - 16} \times 3 = 7.88 \).
Mean.
Class mark \( x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \).
| Class | \( f_i \) | \( x_i \) | \( f_ix_i \) |
|---|---|---|---|
| 1 - 4 | 6 | 2.5 | 15 |
| 4 - 7 | 30 | 5.5 | 165 |
| 7 - 10 | 40 | 8.5 | 340 |
| 10 - 13 | 16 | 11.5 | 184 |
| 13 - 16 | 4 | 14.5 | 58 |
| 16 - 19 | 4 | 17.5 | 70 |
| Total | \( \sum f_i = 100 \) | \( \sum f_ix_i = 832 \) |
Mean \( \bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{832}{100} = 8.32 \).
In simple words: Find median, mode and mean one after the other from the same table.
Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.
| 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 |
Answer:
| Class | Frequency | Cumulative frequency |
|---|---|---|
| 40 - 45 | 2 | 2 |
| 45 - 50 | 3 | 5 |
| 50 - 55 | 8 | 13 |
| 55 - 60 | 6 | 19 |
| 60 - 65 | 6 | 25 |
| 65 - 70 | 3 | 28 |
| 70 - 75 | 2 | 30 |
\( N = 30 \), so \( \frac{N}{2} = 15 \). The cumulative frequency just greater than this is 19, so the median class is 55 - 60.
\( l = 55,\ c.f. = 13,\ f = 6,\ h = 5 \).
Median \( = l + \frac{\frac{N}{2} - c.f.}{f} \times h = 55 + \frac{15 - 13}{6} \times 5 = 56.67 \).
Therefore the median weight of the students is about 56.67 kg.
In simple words: Median class 55 - 60 contains the 15th observation.
Teacher's Note:
1. Build the cumulative frequency column first and find N/2 before choosing the median class.
2. The value of c.f. in the formula belongs to the class before the median class.
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Download Practice Assignments: Class 10 Mathematics Chapter 13 Statistics
Class 10 Mathematics Chapter 13 Statistics Printable Assignments
Review targeted chapter assignments for Class 10 Mathematics Chapter 13 Statistics. Built according to official CBSE guidelines, these downloadable problem sets help students build accuracy and prepare effectively for school tests.
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