STATISTICS-Exercise 13.3

Q1. 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	4

Solution

Lets us put the data in a frequency distribution table

Class Interval	\(f_i\)	\(cf\)	\(x_i\)	\(d_i\)	\(f_id_i\)
65-85	4	4	75	-60	-240
85-105	5	9	95	-40	-200
105-125	13 \((f_0)\)	22	115	-20	-260
125-145	20 \((f_1)\)	42	135	0	0
145-165	14 \((f_2)\)	56	155	20	280
165-185	8	64	175	40	320
185-205	4	68	195	60	240
Total	\(\sum{f_i}=68\)	\(cf=68\)		\(\sum{f_id_i}=140\)

Calculation of Mode
Modal Class = 125-145 (class having maximum frequency)
\(h\) = upper class limit-lower class limit = 85-65 = 20

\(\begin{aligned}l&=125\\ h&=2\\ f_{0}&=13\\ f_{1}&=20\\ f_{2}&=14\end{aligned}\)
Substituting Values in Mode Formula $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-}f_{2}\right) h\\\\ &=125+\left( \dfrac{20-13}{2\times 2-13-14}\right) h\\\\ &=125+\left( \dfrac{7}{40-27}\right) \times 20\\\\ &=125+\left( \dfrac{7\times 20}{13}\right) \\\\ &=125+\dfrac{140}{13}\\\\ &=135.76\end{aligned}$$

Calculation of Mean
Assumed mean \(a\) = 135
Substituting value in Mean Formula

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}d_{i}}{\sum f_{i}}\right) \\\\ &=135+\left( \dfrac{140}{68}\right) \\\\ &=135+2.05\\\\ &=737.05\end{aligned}$$

Calculation of Median
\(\frac{n}{2}\times 68 = 34\)
Median class = 125-145 (34 lies in this class)
\(cf = 22\) (Cumulative frequency of preceding class)
\(f\) = frequency of median class \(h\) - class size

\(\begin{aligned}\dfrac{n}{2}&=34\\ l&=125\\ cf&=22\\ h&=20\\ f&=20\end{aligned}\) $$\begin{aligned}\text{Median}&=l+\left( \dfrac{\dfrac{n}{2}-cf}{t}\right) \times h\\\\ &=125+\left( \dfrac{34-22}{20}\right) \times 20\\\\ &=125+\left( \dfrac{12\times 20}{20}\right) \\\\ &=125+12\\\\ &=137\end{aligned}$$

Median = 137, Mode = 135.76, Mean = 137.05

Q2. 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

Solution

Let us put data in a frequency table

Class interval	Frequency \(f\)	\(cf\)
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	\(45+x+y\)
Total	60	\(45+x+y\)

Median
$$\begin{align}n&=60\\ \dfrac{n}{2}&=\dfrac{60}{2}\\ &=30\\ 45+x+y&=60\\ x+y&=60-45\\ x+y&=15\tag{1}\end{align}$$

Median 28.5 lies in interval 20-30, therefore median class is 20-30
class size \((h)\) = 10

\(\begin{aligned}l&=20\\ h&=10\\ cf&=5+x\\ f&=0\end{aligned}\)

$$\begin{aligned}\text{Median}&=l+\left( \dfrac{n/2-cf}{f}\right) \times h\\\\ 28.5&=20+\left[ \dfrac{30-\left( 5+x\right) }{20}\right] \times 10\\\\ 28.5&=20+\left[ \dfrac{25-x}{2}\right] \\\\ 28.5-20&=\dfrac{25-x}{2}\\\\ 8.5\times 2&=25-x\\\\ 17&=25-x\\\\ x&=25-17\\\\ x&=8\end{aligned}$$

Substituting \(x=8\) in equation-(1)

$$\begin{aligned}x+y&=15\\\\ y&=15-x\\\\ &=15-8\\\\ &=7\\\\ \end{aligned}$$

\(x=8,\ y=7\)

Q3. 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.

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

Solution

Lets us put the data in frequency table

Class Interval	\(f\)	\(cf\)
Below 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

$$\begin{aligned}\dfrac{n}{2}&=\dfrac{100}{2}\\ &=50\end{aligned}$$

50 lies in class interval of 35-40
therefore median class = 35-40
Class size \((h)\) = 5

\(\begin{aligned}l&=35\\ h&=5\\ cf&=45\\ f&=33\end{aligned}\) $$\begin{aligned}\text{Median}&=l+\left( \dfrac{n/2-cf}{f}\right) \times h\\\\ &=35+\left( \dfrac{50-45}{33}\right) \times 5\\\\ &=35+\left( \dfrac{5\times 5}{33}\right) \\\\ &=35+\dfrac{25}{33}\\\\ &=35+0.76\\\\ &=35.76\end{aligned}$$

Median age of policy holders is 35.76 years

Q4. 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.

Length (in mm)	Number of leaves
116-126	3
127-135	5
136-144	9
145-153	12
154-162	5
163-171	4
172-180	2

Solution

Class Interval is not continuos, therefore to make it continuous we will subtract 0.5 from lower limit and add 0.5 to upper limit Lets put data in frequency table

Class Interval	\(f\)	\(cf\)
115.5-126.5	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

$$\begin{aligned}\dfrac{n}{2}&=\dfrac{40}{2}\\ &=20\end{aligned}$$

20 lies in the interval 144.5- 153.5
therefore median class = 144.5-153.5
Class size \((h)\) = 9

\(\begin{aligned}l&=144.5\\ h&=9\\ cf&=17\\ f&=12\end{aligned}\) $$\begin{aligned}\text{Median}&=l+\left( \dfrac{n/2-cf}{f}\right) h\\\\ &=144.5+\left( \dfrac{20-17}{12}\right) \times 9\\\\ &=144.5+\left( \dfrac{3}{12}\times 9\right) \\\\ &=144.5+\dfrac{9}{4}\\\\ &=144.5+2.25\\\\ &=146.75\end{aligned}$$

Q5. The following table gives the distribution of the life time of 400 neon lamps :

Runs score	Number of batsmen
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.

Solution

Class Interval	\(f\)	\(cf\)
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

$$\begin{aligned}\dfrac{n}{2}=\dfrac{400}{2}\\ =200\end{aligned}$$

200 lies in 3000-3500 class interval
therefore Median class is 3000-3500
Class size \((h)\) = 3500-3000 = 500

\(\begin{aligned}l&=3000\\ h&=500\\ cf&=130\\ f&=86\end{aligned}\) $$\begin{aligned}\text{Median}&=l+\left( \dfrac{n/2-cf}{f}\right) h\\\\ &=3000+\left( \dfrac{200-130}{86}\right) 500\\\\ &=3000+\dfrac{70}{86}\times 500\\\\ &=3000+406.98\\\\ &=3406.98\end{aligned}$$

Median life time of lamp is 3406.98 hours

Q6. 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.

Solution

Lets put this data in frequency table

Class Interval	\(f_i\)	\(cf\)	\(x_i\)	\(d_i=a-x_i\)	\(f_id_i\)
1-4	6	6	2.5	-6	-36
4-7	30	36	5.5	-3	-90
7-10	40	76	8.5	0	0
10-13	16	92	11.5	3	48
13-16	4	96	14.5	6	24
16-19	4	100	15.5	9	36
Total	\(\sum{f_i}=100\)			\(\sum{f_id_i}=-18\)

Calculation of Median

$$\begin{aligned}\dfrac{n}{2}&=\dfrac{100}{2}\\ &=50\end{aligned}$$

50 lies in the interval of 7-10, hence median class is 7-10
class size \((h)\) = 3

\(\begin{aligned}l&=7\\ h&=3\\ cf&=36\\ f&=40\end{aligned}\) $$\begin{aligned}\text{Median}&=l+\left( \dfrac{n/2-cf}{f}\right) h\\\\ &=7+\left( \dfrac{50-36}{40}\right) 3\\\\ &=7+\dfrac{14}{40}\times 3\\\\ &=7+\dfrac{42}{40}\\\\ &=7+1.05\\\\ &=8.05\end{aligned}$$

Calculation of Mean

Assumed mean \(a\) = 8.5

$$\begin{aligned}\overline{x}&=a+\left( \dfrac{\sum f_{i}d_{i}}{\sum fi}\right) \\ &=8.5+\left( \dfrac{-18}{10}\right) \\ &=8.5-0.18\\ &=8.32\end{aligned}$$

Calculation of Mode

Class interval 7 - 10 is modal class as highest frequency lies in this class
class size \((h)\) = 3

\(\begin{aligned}l&=7\\ f_{0}&=30\\ f_{1}&=40\\ f_{2}&=16\\ h&=3\end{aligned}\) $$\begin{aligned}\text{MODE}&=l+\left( \dfrac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right) \times h\\\\ &=7+\left( \dfrac{40-30}{40\times 2-30-16}\right) \times 3\\\\ &=7+\dfrac{10\times 3}{80-46}\\\\ &=7+\dfrac{30}{34}\\\\ &=7+0.88\\\\ &=7.88\end{aligned}$$

Q7. 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

Solution

Lets put these data in frequency table

Class Interval	\(f\)	\(cf\)
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

$$\begin{aligned}\dfrac{n}{2}&=\dfrac{30}{2}\\ &=15\end{aligned}$$

15 lies in iterral 55-60, it is the median class
class size \((h)\) = 5

\(\begin{aligned}l&=55\\ h&=5\\ cf&=13\\ f&=6\end{aligned}\) $$\begin{aligned}\text{Median}&=l+\left( \dfrac{n/2-cf}{f}\right) h\\\\ &=55+\left( \dfrac{15-13}{6}\right) \times 5\\\\ &=55+\dfrac{2\times 5}{6}\\\\ &=55+1.67\\\\ &=56.67\end{aligned}$$

Median weight is 56.67 kg