Mathematics solution NCERT
Class 10 - Chapter 13: Statistics
Exercise 13.3
Question 1 (Part 1: Mean)
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the mean of the data.
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take the assumed mean
$$ A=135, $$and class width
$$ h=20. $$The class marks are
$$ 75,\;95,\;115,\;135,\;155,\;175,\;195. $$| Monthly Consumption (Units) | Frequency (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 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 | 68 | — | — | 7 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the values.
$$ \bar{x} = 135 + 20\left(\frac{7}{68}\right) $$ $$ = 135+\frac{140}{68} $$ $$ = 135+2.0588 $$ $$ = 137.0588 $$Therefore,
$$ \boxed{\bar{x}\approx137.06\text{ units}} $$Answer:
$$ \boxed{\text{Mean monthly consumption}=137.06\text{ units}} $$Question 1 (Part 2: Median)
Find the median monthly consumption of electricity.
Solution:
First, prepare the cumulative frequency (CF) table.
| Monthly Consumption (Units) | Frequency (f) | Cumulative Frequency (CF) |
|---|---|---|
| 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 |
Total number of consumers is
$$ N=68. $$Therefore,
$$ \frac{N}{2}=\frac{68}{2}=34. $$The cumulative frequency just greater than 34 is 42. Hence, the median class is
$$ 125-145. $$For the median class,
$$ l=125,\qquad h=20, $$ $$ f=20,\qquad cf=22,\qquad N=68. $$Using the formula for the median of grouped data,
$$ \text{Median} = l+ \left( \frac{\frac{N}{2}-cf}{f} \right)h $$Substitute the given values.
$$ \text{Median} = 125+ \left( \frac{34-22}{20} \right)\times20 $$ $$ = 125+\frac{12}{20}\times20 $$ $$ = 125+12 $$ $$ = 137 $$Answer:
$$ \boxed{\text{Median monthly consumption}=137\text{ units}} $$Question 1 (Part 3: Mode and Comparison)
Find the mode of the data and compare the mean, median and mode.
Step 1: Find the Mode
The highest frequency is
$$ 20. $$Hence, the modal class is
$$ 125-145. $$For the modal class,
$$ l=125,\qquad h=20, $$ $$ f_1=20,\qquad f_0=13,\qquad f_2=14. $$Using the formula for the mode of grouped data,
$$ \text{Mode} = l+ \left( \frac{f_1-f_0} {2f_1-f_0-f_2} \right)h $$Substitute the given values.
$$ \text{Mode} = 125+ \left( \frac{20-13} {2(20)-13-14} \right)\times20 $$ $$ = 125+ \left( \frac{7} {40-27} \right)\times20 $$ $$ = 125+\frac{140}{13} $$ $$ = 125+10.77 $$ $$ = 135.77 $$Mode
$$ \boxed{\text{Mode}\approx135.77\text{ units}} $$Comparison of Mean, Median and Mode
From the previous parts, we have:
| Measure of Central Tendency | Value (Units) |
|---|---|
| Mean | 137.06 |
| Median | 137 |
| Mode | 135.77 |
The three measures are very close to one another.
$$ \text{Mean}\approx137.06 $$ $$ \text{Median}=137 $$ $$ \text{Mode}\approx135.77 $$This shows that the distribution is nearly symmetrical. The monthly electricity consumption of most consumers is concentrated around 136–137 units.
Answer:
Mode:
$$ \boxed{135.77\text{ units}} $$Comparison:
$$ \boxed{ \text{Mean}\approx137.06,\; \text{Median}=137,\; \text{Mode}\approx135.77 } $$Interpretation: Since the mean, median and mode are very close to each other, the data is nearly symmetric. The average monthly electricity consumption of the consumers is about 137 units, and the most common consumption is about 136 units.
Question 2
If the median of the distribution given below is 28.5, find the values of x and y.
Solution:
The given frequency distribution is:
| Class Interval | Frequency |
|---|---|
| 0–10 | 5 |
| 10–20 | x |
| 20–30 | 20 |
| 30–40 | 15 |
| 40–50 | y |
| 50–60 | 5 |
| Total | 60 |
Step 1: Form the first equation.
Total frequency is
$$ 60. $$Therefore,
$$ 5+x+20+15+y+5=60 $$ $$ 45+x+y=60 $$ $$ \boxed{x+y=15} \qquad\cdots(1) $$Step 2: Identify the median class.
Total frequency
$$ N=60 $$Hence,
$$ \frac{N}{2}=30. $$Since the median is
$$ 28.5, $$the median class is
$$ 20-30. $$For the median class,
$$ l=20, $$ $$ h=10, $$ $$ f=20, $$ $$ cf=5+x. $$Step 3: Apply the median formula.
$$ \text{Median} = l+ \left( \frac{\frac{N}{2}-cf}{f} \right)h $$Substitute the given values.
$$ 28.5 = 20+ \left( \frac{30-(5+x)}{20} \right)\times10 $$ $$ 28.5 = 20+ \frac{25-x}{2} $$Subtract 20 from both sides.
$$ 8.5=\frac{25-x}{2} $$Multiply both sides by 2.
$$ 17=25-x $$ $$ x=8 $$Step 4: Find the value of y.
Using equation (1),
$$ x+y=15 $$ $$ 8+y=15 $$ $$ y=7 $$Verification:
The cumulative frequencies are:
| Class Interval | Frequency | Cumulative Frequency |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 8 | 13 |
| 20–30 | 20 | 33 |
| 30–40 | 15 | 48 |
| 40–50 | 7 | 55 |
| 50–60 | 5 | 60 |
Since
$$ 30 $$lies between cumulative frequencies
$$ 13 \quad\text{and}\quad 33, $$the median class is correctly identified as
$$ 20-30. $$Using the values
$$ x=8,\qquad y=7, $$the median obtained is
$$ 28.5. $$Hence, the values are correct.
Answer:
$$ \boxed{x=8,\qquad y=7} $$Question 3
A life insurance agent found the following data for the 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 years.
Solution:
The given data is a less than cumulative frequency distribution. First, convert it into a simple frequency distribution.
| Age (Years) | Less Than Cumulative Frequency | Frequency (f) | Cumulative Frequency (CF) |
|---|---|---|---|
| 18–20 | 2 | 2 | 2 |
| 20–25 | 6 | 6−2 = 4 | 6 |
| 25–30 | 24 | 24−6 = 18 | 24 |
| 30–35 | 45 | 45−24 = 21 | 45 |
| 35–40 | 78 | 78−45 = 33 | 78 |
| 40–45 | 89 | 89−78 = 11 | 89 |
| 45–50 | 92 | 92−89 = 3 | 92 |
| 50–55 | 98 | 98−92 = 6 | 98 |
| 55–60 | 100 | 100−98 = 2 | 100 |
| Total | 100 | 100 | — |
Step 1: Find the median class.
Total number of policy holders is
$$ N=100 $$Therefore,
$$ \frac{N}{2}=\frac{100}{2}=50 $$The cumulative frequency just greater than 50 is
$$ 78. $$Hence, the median class is
$$ 35-40. $$For the median class,
$$ l=35,\qquad h=5, $$ $$ f=33,\qquad cf=45. $$Step 2: Apply the median formula.
$$ \text{Median} = l+ \left( \frac{\frac{N}{2}-cf}{f} \right)h $$Substitute the values.
$$ \text{Median} = 35+ \left( \frac{50-45}{33} \right)\times5 $$ $$ = 35+\frac{25}{33} $$ $$ = 35+0.758 $$ $$ = 35.758 $$Answer:
$$ \boxed{\text{Median Age}\approx35.76\text{ years}} $$Therefore, the median age of the policy holders is approximately
$$ \boxed{35.8\text{ years}.} $$Question 4
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. Find the median length of the leaves.
Solution:
Since the measurements are recorded to the nearest millimetre, first convert the given classes into continuous class intervals.
| Length (mm) | Continuous Class | Frequency (f) | Cumulative Frequency (CF) |
|---|---|---|---|
| 118–126 | 117.5–126.5 | 3 | 3 |
| 127–135 | 126.5–135.5 | 5 | 8 |
| 136–144 | 135.5–144.5 | 9 | 17 |
| 145–153 | 144.5–153.5 | 12 | 29 |
| 154–162 | 153.5–162.5 | 5 | 34 |
| 163–171 | 162.5–171.5 | 4 | 38 |
| 172–180 | 171.5–180.5 | 2 | 40 |
| Total | — | 40 | — |
Step 1: Find the median class.
Total number of leaves is
$$ N=40 $$Therefore,
$$ \frac{N}{2}=\frac{40}{2}=20 $$The cumulative frequency just greater than
$$ 20 $$is
$$ 29. $$Hence, the median class is
$$ 144.5-153.5. $$For the median class,
$$ l=144.5,\qquad h=9, $$ $$ f=12,\qquad cf=17. $$Step 2: Apply the median formula.
$$ \text{Median} = l+ \left( \frac{\frac{N}{2}-cf}{f} \right)h $$Substitute the values.
$$ \text{Median} = 144.5+ \left( \frac{20-17}{12} \right)\times9 $$ $$ = 144.5+ \frac{27}{12} $$ $$ = 144.5+2.25 $$ $$ = 146.75 $$Answer:
$$ \boxed{\text{Median length}=146.75\text{ mm}} $$Therefore, the median length of the leaves is
$$ \boxed{146.75\text{ mm}.} $$Question 5
The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.
Solution:
First, prepare the cumulative frequency (CF) table.
| Life Time (Hours) | Frequency (f) | Cumulative Frequency (CF) |
|---|---|---|
| 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 |
| Total | 400 | — |
Step 1: Find the median class.
Total number of lamps is
$$ N=400 $$Therefore,
$$ \frac{N}{2}=\frac{400}{2}=200 $$The cumulative frequency just greater than
$$ 200 $$is
$$ 216. $$Hence, the median class is
$$ 3000-3500. $$For the median class,
$$ l=3000,\qquad h=500, $$ $$ f=86,\qquad cf=130. $$Step 2: Apply the median formula.
$$ \text{Median} = l+ \left( \frac{\frac{N}{2}-cf}{f} \right)h $$Substitute the values.
$$ \text{Median} = 3000+ \left( \frac{200-130}{86} \right)\times500 $$ $$ = 3000+ \frac{70}{86}\times500 $$ $$ = 3000+406.98 $$ $$ = 3406.98 $$Answer:
$$ \boxed{\text{Median life time}=3406.98\text{ hours}} $$Therefore, the median life time of a neon lamp is approximately
$$ \boxed{3407\text{ hours}.} $$Question 6 (Part 1: Median)
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the surnames was obtained. Determine the median number of letters in the surnames.
Solution:
First, prepare the cumulative frequency (CF) table.
| Number of Letters | Frequency (f) | Cumulative Frequency (CF) |
|---|---|---|
| 1–4 | 6 | 6 |
| 4–7 | 30 | 36 |
| 7–10 | 40 | 76 |
| 10–13 | 16 | 92 |
| 13–16 | 4 | 96 |
| 16–19 | 4 | 100 |
| Total | 100 | — |
Step 1: Find the median class.
Total number of surnames is
$$ N=100 $$Therefore,
$$ \frac{N}{2}=\frac{100}{2}=50 $$The cumulative frequency just greater than
$$ 50 $$is
$$ 76. $$Hence, the median class is
$$ 7-10. $$For the median class,
$$ l=7,\qquad h=3, $$ $$ f=40,\qquad cf=36. $$Step 2: Apply the median formula.
$$ \text{Median} = l+ \left( \frac{\frac{N}{2}-cf}{f} \right)h $$Substitute the values.
$$ \text{Median} = 7+ \left( \frac{50-36}{40} \right)\times3 $$ $$ = 7+\frac{14}{40}\times3 $$ $$ = 7+1.05 $$ $$ = 8.05 $$Answer:
$$ \boxed{\text{Median number of letters}=8.05} $$Therefore, the median number of letters in the surnames is approximately
$$ \boxed{8.05.} $$Question 6 (Part 2: Mean)
Find the mean number of letters in the surnames.
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take the assumed mean
$$ A=8.5, $$and class width
$$ h=3. $$The class marks are
$$ 2.5,\;5.5,\;8.5,\;11.5,\;14.5,\;17.5. $$| Number of Letters | Frequency (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 1–4 | 6 | 2.5 | -2 | -12 |
| 4–7 | 30 | 5.5 | -1 | -30 |
| 7–10 | 40 | 8.5 | 0 | 0 |
| 10–13 | 16 | 11.5 | 1 | 16 |
| 13–16 | 4 | 14.5 | 2 | 8 |
| 16–19 | 4 | 17.5 | 3 | 12 |
| Total | 100 | — | — | -6 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the values.
$$ \bar{x} = 8.5+ 3\left(\frac{-6}{100}\right) $$ $$ = 8.5-0.18 $$ $$ = 8.32 $$Answer:
$$ \boxed{\text{Mean number of letters}=8.32} $$Question 6 (Part 3: Mode)
Find the modal size of the surnames.
Solution:
The highest frequency is
$$ 40. $$Hence, the modal class is
$$ 7-10. $$| Number of Letters | Frequency (f) |
|---|---|
| 1–4 | 6 |
| 4–7 | 30 |
| 7–10 | 40 |
| 10–13 | 16 |
| 13–16 | 4 |
| 16–19 | 4 |
For the modal class,
$$ l=7,\qquad h=3, $$ $$ f_1=40,\qquad f_0=30,\qquad f_2=16. $$Using the formula for the mode of grouped data,
$$ \text{Mode} = l+ \left( \frac{f_1-f_0} {2f_1-f_0-f_2} \right)h $$Substitute the given values.
$$ \text{Mode} = 7+ \left( \frac{40-30} {2(40)-30-16} \right)\times3 $$ $$ = 7+ \left( \frac{10} {80-46} \right)\times3 $$ $$ = 7+\frac{30}{34} $$ $$ = 7+0.882 $$ $$ = 7.882 $$Mode
$$ \boxed{\text{Mode}\approx7.88} $$Final Results
| Measure of Central Tendency | Value |
|---|---|
| Median | 8.05 |
| Mean | 8.32 |
| Mode | 7.88 |
Interpretation:
The three measures of central tendency are very close to one another:
$$ \boxed{\text{Mode}\approx7.88,\qquad \text{Median}=8.05,\qquad \text{Mean}=8.32} $$This indicates that the distribution of the number of letters in the surnames is nearly symmetrical. On average, most surnames contain about 8 letters.
Answer:
Modal size of the surnames:
$$ \boxed{7.88\text{ letters}\approx8\text{ letters}} $$Question 7
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Solution:
First, prepare the cumulative frequency (CF) table.
| Weight (kg) | Frequency (f) | Cumulative Frequency (CF) |
|---|---|---|
| 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 |
| Total | 30 | — |
Step 1: Find the median class.
Total number of students is
$$ N=30 $$Therefore,
$$ \frac{N}{2}=\frac{30}{2}=15 $$The cumulative frequency just greater than
$$ 15 $$is
$$ 19. $$Hence, the median class is
$$ 55-60. $$For the median class,
$$ l=55,\qquad h=5, $$ $$ f=6,\qquad cf=13. $$Step 2: Apply the median formula.
$$ \text{Median} = l+ \left( \frac{\frac{N}{2}-cf}{f} \right)h $$Substitute the values.
$$ \text{Median} = 55+ \left( \frac{15-13}{6} \right)\times5 $$ $$ = 55+\frac{10}{6} $$ $$ = 55+1.67 $$ $$ = 56.67 $$Answer:
$$ \boxed{\text{Median weight}=56.67\text{ kg}} $$Therefore, the median weight of the students is approximately
$$ \boxed{56.67\text{ kg}.} $$