Mathematics solution NCERT
Class 10 - Chapter 13: Statistics
Exercise 13.2
Question 1
The following table shows the ages of the patients admitted in a hospital during a year. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
Step 1: Find the Mean
Since the class intervals are equal, we use the Direct Method.
| Age (Years) | Frequency (f) | Class Mark (x) | f × x |
|---|---|---|---|
| 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 | 80 | — | 2830 |
Using the Direct Method,
$$ \bar{x}=\frac{\sum fx}{\sum f} $$ $$ \bar{x}=\frac{2830}{80} $$ $$ \boxed{\bar{x}=35.375\text{ years}} $$Step 2: Find the Mode
The highest frequency is
$$ 23, $$so the modal class is
$$ 35-45. $$Here,
$$ l=35,\qquad h=10, $$ $$ f_1=23,\qquad f_0=21,\qquad f_2=14. $$Using the formula
$$ \text{Mode} = l+ \left( \frac{f_1-f_0} {2f_1-f_0-f_2} \right)h $$Substitute the values.
$$ \text{Mode} = 35+ \left( \frac{23-21} {2(23)-21-14} \right)\times10 $$ $$ = 35+ \left( \frac{2}{46-35} \right)\times10 $$ $$ = 35+\frac{20}{11} $$ $$ = 35+1.818 $$ $$ \boxed{\text{Mode}\approx36.82\text{ years}} $$Comparison and Interpretation
The two measures of central tendency are:
$$ \boxed{\text{Mean}=35.38\text{ years}} $$ $$ \boxed{\text{Mode}=36.82\text{ years}} $$The mean and the mode are very close to each other. This indicates that most patients admitted to the hospital were around 35–37 years of age. The age distribution is fairly balanced, and there is no large variation in the ages of the patients.
Answer:
$$ \boxed{\text{Mean}=35.38\text{ years}} $$ $$ \boxed{\text{Mode}\approx36.82\text{ years}} $$Interpretation: Most patients admitted during the year were approximately 36 years old, and the average age of all patients was about 35 years.
Question 2
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components. Determine the modal lifetime of the components.
Solution:
The highest frequency is
$$ 61. $$Therefore, the modal class is
$$ 60-80. $$| Lifetime (Hours) | Frequency (f) |
|---|---|
| 0–20 | 10 |
| 20–40 | 35 |
| 40–60 | 52 |
| 60–80 | 61 |
| 80–100 | 38 |
| 100–120 | 29 |
For the modal class,
$$ l=60,\qquad h=20, $$ $$ f_1=61,\qquad f_0=52,\qquad f_2=38. $$Using the formula for 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.
$$ \text{Mode} = 60+ \left( \frac{61-52} {2(61)-52-38} \right)\times20 $$ $$ = 60+ \left( \frac{9} {122-90} \right)\times20 $$ $$ = 60+ \frac{180}{32} $$ $$ = 60+5.625 $$ $$ = 65.625 $$Answer:
$$ \boxed{\text{Modal lifetime}=65.625\text{ hours}} $$Hence, the modal lifetime of the electrical components is approximately
$$ \boxed{65.63\text{ hours}.} $$Question 3
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.
Part A: Find the Modal Monthly Expenditure
Solution:
The highest frequency is
$$ 40. $$Therefore, the modal class is
$$ 1500-2000. $$Here,
$$ l=1500,\qquad h=500, $$ $$ f_1=40,\qquad f_0=24,\qquad f_2=33. $$Using the formula for mode of grouped data,
$$ \text{Mode} = l+ \left( \frac{f_1-f_0} {2f_1-f_0-f_2} \right)h $$Substitute the values.
$$ \text{Mode} = 1500+ \left( \frac{40-24} {2(40)-24-33} \right)\times500 $$ $$ = 1500+ \left( \frac{16} {80-57} \right)\times500 $$ $$ = 1500+ \frac{8000}{23} $$ $$ = 1500+347.83 $$ $$ = 1847.83 $$Modal Monthly Expenditure
$$ \boxed{\text{Mode}\approx₹1847.83} $$Part B: Find the Mean Monthly Expenditure
Since all class intervals are equal, we use the Step-Deviation Method.
Take
$$ A=2750,\qquad h=500. $$| Expenditure (₹) | Frequency (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 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 | 200 | — | — | -35 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the values.
$$ \bar{x} = 2750 + 500\left(\frac{-35}{200}\right) $$ $$ = 2750-87.5 $$ $$ = 2662.5 $$Mean Monthly Expenditure
$$ \boxed{\bar{x}=₹2662.50} $$Answer
Modal Monthly Expenditure:
$$ \boxed{₹1847.83} $$Mean Monthly Expenditure:
$$ \boxed{₹2662.50} $$Question 4
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.
Part A: Mean of the Distribution
Solution:
Since all the class intervals are equal, we use the Step-Deviation Method.
Take
$$ A=32.5,\qquad h=5. $$The class marks are
$$ 17.5,\;22.5,\;27.5,\;32.5,\;37.5,\;42.5,\;47.5,\;52.5. $$| Students per Teacher | Number of States/U.T. (f) | Class Mark (x) | $$ u=\frac{x-A}{h} $$ | fu |
|---|---|---|---|---|
| 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 | 35 | — | — | -23 |
Using the Step-Deviation Method,
$$ \bar{x} = A+h\left(\frac{\sum fu}{\sum f}\right) $$Substitute the values.
$$ \bar{x} = 32.5 + 5\left(\frac{-23}{35}\right) $$ $$ = 32.5-3.286 $$ $$ = 29.214 $$Therefore,
$$ \boxed{\bar{x}\approx29.21} $$Part B: Mode of the Distribution
The highest frequency is
$$ 10. $$Hence, the modal class is
$$ 30-35. $$Here,
$$ l=30,\qquad h=5, $$ $$ f_1=10,\qquad f_0=9,\qquad f_2=3. $$Using the mode formula,
$$ \text{Mode} = l+ \left( \frac{f_1-f_0} {2f_1-f_0-f_2} \right)h $$Substitute the values.
$$ \text{Mode} = 30+ \left( \frac{10-9} {2(10)-9-3} \right)\times5 $$ $$ = 30+ \frac{5}{8} $$ $$ = 30.625 $$Mode
$$ \boxed{\text{Mode}\approx30.63} $$Interpretation
The mean teacher-student ratio is
$$ 29.21, $$while the modal teacher-student ratio is
$$ 30.63. $$Both values are close to each other. This indicates that in most states and Union Territories, there are about 29 to 31 students per teacher. Thus, the distribution is fairly balanced, and the teacher-student ratio in India is generally around 30 students per teacher.
Answer:
Mean teacher-student ratio:
$$ \boxed{29.21} $$Mode:
$$ \boxed{30.63} $$Interpretation: Most states and Union Territories have approximately 30 students per teacher, and the average teacher-student ratio is about 29 students per teacher.
Question 5
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Find the mode of the data.
Solution:
The highest frequency is
$$ 18. $$Hence, the modal class is
$$ 4000-5000. $$| Runs Scored | Number of Batsmen (f) |
|---|---|
| 3000–4000 | 4 |
| 4000–5000 | 18 |
| 5000–6000 | 9 |
| 6000–7000 | 7 |
| 7000–8000 | 6 |
| 8000–9000 | 3 |
| 9000–10000 | 1 |
| 10000–11000 | 1 |
For the modal class,
$$ l=4000,\qquad h=1000, $$ $$ f_1=18,\qquad f_0=4,\qquad f_2=9. $$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.
$$ \text{Mode} = 4000+ \left( \frac{18-4} {2(18)-4-9} \right)\times1000 $$ $$ = 4000+ \left( \frac{14} {36-13} \right)\times1000 $$ $$ = 4000+ \frac{14000}{23} $$ $$ = 4000+608.70 $$ $$ = 4608.70 $$Answer:
$$ \boxed{\text{Mode}\approx4608.70\text{ runs}} $$Hence, the most common number of runs scored by the batsmen is approximately
$$ \boxed{4609\text{ runs}.} $$Question 6
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.
Solution:
The highest frequency is
$$ 20. $$Hence, the modal class is
$$ 40-50. $$| Number of Cars | Frequency (f) |
|---|---|
| 0–10 | 7 |
| 10–20 | 14 |
| 20–30 | 13 |
| 30–40 | 12 |
| 40–50 | 20 |
| 50–60 | 11 |
| 60–70 | 15 |
| 70–80 | 8 |
For the modal class,
$$ l=40,\qquad h=10, $$ $$ f_1=20,\qquad f_0=12,\qquad f_2=11. $$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.
$$ \text{Mode} = 40+ \left( \frac{20-12} {2(20)-12-11} \right)\times10 $$ $$ = 40+ \left( \frac{8} {40-23} \right)\times10 $$ $$ = 40+ \frac{80}{17} $$ $$ = 40+4.706 $$ $$ = 44.706 $$Answer:
$$ \boxed{\text{Mode}\approx44.71} $$Therefore, the most frequently occurring number of cars passing through the road in a 3-minute period is approximately
$$ \boxed{45\text{ cars}.} $$