Q1. 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.

Solution

Lets us put the data in table

Class Interval	\(f_i\)	\(x_i\)	\(d_i\)	\(f_id_i\)
5-15	6	10	-20	-120
15-25	11	20	-10	-110
25-35	21 \((f_0)\)	30	0	0
35-45	23 \((f_1)\)	40	10	230
45-55	14 \((f_2)\)	50	20	280
55-65	5	60	30	150
Total	\(\sum{f_i=80}\)		\(\sum{f_id_i=430}\)

Modal Class 35-45 (class with highest frequency)
class interval (h) = Upper class limit-lower class limit = 15-5- 10

$$\begin{aligned}l=35\\ f_{1}=23\\ f_{0}=21\\ f_{2}=14\\ h=10\\ \text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) h\\\\ &=35+\left( \dfrac{23-21}{2\times 23-21-14}\right) 10\\\\ &=35+\left( \dfrac{2}{46-35}\right) \times 10\\\\ &=35+\dfrac{2}{11}\times 10\\\\ &=35+1.8\\\\ &=36.8\end{aligned}$$

Assumed mean = 30
Substitute the values

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\right) \\\\ &=30+\left( \dfrac{430}{80}\right) \\\\ &=30+\dfrac{43}{8}\\\\ &=35.37\end{aligned}$$ Mode = 36.8 Mean = 35.37

Maximum Number of patients admitted in the hospital are of age 36.8, while on an average the age of patient admitted to the hospital is 35.37 years

Q2. 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 \(\tiny\text{(Modal Class)}\)	80-100	100-120
Frequency	10	35	52 \((f_0)\)	61 \((f_1)\)	38 \((f_2)\)	29

Determine the modal lifetimes of the components.

Solution

Modal Class = 60-80 (class having Maxium frequency)
class interval (h) = Upper class limit-lower class limit = 20-0 = 20

\(\begin{aligned}l=60\\ h=20\\ f_{0}=52\\ f_{1}=61\\ f_{2}=38\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) h\\\\ &=60+\left( \dfrac{61-52}{2\times 61-52-38}\right) \times 20\\\\ &=60+\left( \dfrac{9}{122-90}\right) \times 20\\\\ &=60+\left( \dfrac{9}{32}\right) \times 20\\\\ &=65.625\end{aligned}$$

Modal lifetime of components = 65.625 hours

Q3. 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 ₹)	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

Solution

Lets us put the data in table

Class Interval	\(f_i\)	\(x_i\)	\(d_i=x_i-a\)	\(u_i=\frac{d_i}{h}\)	\(f_iu_i\)
1000-1500	24 \((f_0)\)	1250	-1500	-3	-72
1500-2000	40 \((f_1)\)	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750	0	0	0
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	3750	1500	3	48
4500-5000	7	4250	2000	4	28
Total	\(\sum{f_i}=(200)\)		\(\sum{f_iu_i}=(-35)\)

Modal Class = 1500-2000 (Class having Maximum frequency)
Class interval (h) = Upper class limit- lower class limit = 1500-1000 = 500

\(\begin{aligned}l&=1500\\ h&=500\\ f_{0}&=24\\ f_{1}&=40\\ f_{2}&=33\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) h\\\\ &=1500+\left( \dfrac{40-24}{2\times 40-24-33}\right) \times 500\\\\ &=1500+\left( \dfrac{16}{80-57}\right) 500\\\\ &=1500+\dfrac{16\times 500}{23}\\\\ &=1500+347.83\\\\ &=1847.83\end{aligned}$$

Assumed mean = 2750
Class interval (h) = 500
Substituting values in Mean formula

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}u_{i}}{\sum f_{i}}\right) \times h\\\\ &=2750+\left( \dfrac{-35}{200}\right) 500\\\\ &=2750-\dfrac{35\times 5}{2}\\\\ &=2750-87.5\\\\ &=2662.5\end{aligned}$$

Modal Month y Expense = 1847.83
Mean Monthly Expense = 2662.50

Q4. 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

Solution

Lets put data in table

Class Interval	\(f_i\)	\(x_i\)	\(d_i\)	\(f_id_i\)
10-20	3	17.5	-15	-45
20-25	8	22.5	-10	-80
25-30	9 \((f_0)\)	27.5	-5	-45
30-35	10 \((f_1)\)	32.5	0	0
35-40	3 \((f_2)\)	37.5	5	15
40-45	0	42.5	10	0
45-50	0	47.5	15	0
50-55	2	52.5	20	40
Total	\(\sum{f_i}=35\)		\(\sum{f_id_i}=-115\)

Modal Class - 30-35 (Class having maximum frequency)
class interval h = Upper Class limit-lower class limit = 20-15 = 5

\(\begin{aligned}l&=30\\ h&=5\\ f_{0}&=9\\ f_{1}&=10\\ f_{2}&=3\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) \times h\\\\ &=30+\left( \dfrac{10-9}{2\times 10-9-3}\right) 5\\\\ &=30+\left( \dfrac{1}{20-12}\right) 5\\\\ &=30+\dfrac{5}{8}\\\\ &=30.6\end{aligned}$$

Assumed mean = 32.5
Substitute values in formula

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\right) \\\\ &=32.5+\left( \dfrac{-115}{35}\right) \\\\ &=32.5-\dfrac{115}{35}\\\\ &=32.5-\dfrac{23}{7}\\\\ &=29.2\end{aligned}$$

Mode=30.6
Mean = 29.2

Q5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs score	Number of batsmen
3000-4000	4 \((f_0)\)
4000-5000	18 \((f_1)\)
5000-6000	9 \((f_2)\)
6000-7000	7
7000-8000	6
8000-9000	3
9000-10000	2
10000-11000	1

Find the mode of the data.

Solution

Modal class = 4000-5000 (having maximum frequency)
h = 5000-4000 = 1000

\(\begin{aligned}l&=4000\\ h&=1000\\ f_{0}&=4\\ f_{1}&=18\\ f_{2}&=9\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) \times h\\\\ &=4000t\left( \dfrac{18-4}{2\times 18-4-9}\right) \times 1000\\\\ &=4000+\dfrac{14\times 1000}{36-13}\\\\ &=4000+\dfrac{14\times 1000}{23}\\\\ &=4000+608.7\\\\ &=4608.7\end{aligned}$$

Mode = 4608.7

Q6. 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 \((f_0)\)	20 \((f_1)\)	11 \((f_2)\)	15	8

Solution

Modal class = 40-50 (class having maximum frequency)
h = upper Class limit-lower Class limit = 50-40 = 10

\(\begin{aligned}l&=40\\ h&=10\\ f_{0}&=12\\ f_{1}&=20\\ f_{2}&=11\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) \times 10\\\\ &=40+\left( \dfrac{20-12}{2\times 20-12-11}\right) \times 10\\\\ &=40+\left( \dfrac{8}{40-23}\right) \times 10\\\\ &=40+\dfrac{8\times 10}{17}\\\\ &=40+4.7\\\\ &=44.7\end{aligned}$$

Mode = 44.7