Chapter 13 Statistics Exercise-03
EXERCISE 13.3
1. 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 |
| $165-185$ | 8 |
| $185-205$ | 4 |
Show Answer
Solution
To find the class marks, the following relation is used.
Class mark $=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Taking 135 as assumed mean (a), $d_i, u_i, u_i$ are calculated according to step deviation method as follows.
| Monthly consumption (in units) |
Number of consumers $(\boldsymbol{{}f} _{\boldsymbol{{}i}})$ |
$\boldsymbol{{}x} _{\boldsymbol{{}i}}$ class mark |
$\boldsymbol{{}d}_i=\boldsymbol{{}x}_i-$ $\mathbf{1 3 5}$ |
$\boldsymbol{{}u} _{\boldsymbol{{}i}}=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{2 0}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}u} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $65-85$ | 4 | 75 | -60 | -3 | -12 |
| $85-105$ | 5 | 95 | -40 | -2 | -10 |
| $105-125$ | 13 | 115 | -20 | -1 | -13 |
| $125-145$ | 20 | 135 | 0 | 0 | 0 |
| $145-165$ | 14 | 155 | 20 | 1 | 14 |
| $165-185$ | 8 | 175 | 40 | 2 | 16 |
| $185-205$ | 4 | 195 | 60 | 3 | 12 |
| Total | 68 | 7 |
From the table, we obtain
$\sum f_i u_i=7$ $\sum f_i=68$
Class size $(h)=20$
Mean, $\bar{{}x}=a+(\dfrac{\sum f_i u_i}{\sum f_i}) \times h$
$ \begin{aligned} & =135+\dfrac{7}{68} \times 20 \\ & =135+\dfrac{140}{68} \\ & =137.058 \end{aligned} $
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 - 145.
Modal class $=125-145$
Lower limit ( $($ ) of modal class $=125$
Class size $(h)=20$
Frequency $(f_1)$ of modal class $=20$
Frequency $(f_0)$ of class preceding modal class $=13$
Frequency $(f_2)$ of class succeeding the modal cla
2. 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 |
Show Answer
Solution
The cumulative frequency for the given data is calculated as follows.
| 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 | $45+x+y$ |
| Total $(n)$ | 60 |
From the table, it can be observed that $n=60$
$45+x+y=60$
$x+y=15(1)$
Median of the data is given as 28.5 which lies in interval 20 - 30 .
Therefore, median class $=20-30$
Lower limit ( $($ ) of median class $=20$
Cumulative frequency ( $c f$ ) of class preceding the median class $=5+x$
Frequency $(f)$ of median class $=20$
Class size $(h)=10$
Median $=l+(\dfrac{\dfrac{n}{2}-c f}{f}) \times h$
$28.5=20+[\dfrac{\dfrac{60}{2}-(5+x)}{20}] \times 10$
$8.5=(\dfrac{25-x}{2})$
$17=25-x$
$x=8$
From equation (1),
$8+y=15$
$y=7$
Hence, the values of $x$ and $y$ are 8 and 7 respectively.
3. 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 |
Show Answer
Solution
Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.
| Age (in years) | Number of policy holders $(\boldsymbol{{}f} _{\boldsymbol{{}i}})$ | Cumulative frequency $(\boldsymbol{{}c} \boldsymbol{{})})$ |
|---|---|---|
| $18-20$ | 2 | 2 |
| $20-25$ | $6-2=4$ | 6 |
| $25-30$ | $24-6=18$ | 24 |
| $30-35$ | $45-24=21$ | 45 |
| $35-40$ | $78-45=33$ | 78 |
| $40-45$ | $89-78=11$ | 99 |
| $45-50$ | $92-89=3$ | 98 |
| $50-55$ | $98-92=6$ | 100 |
| $55-60$ | $100-98=2$ | |
| Total $(n)$ | 92 |
From the table, it can be observed that $n=100$.
Cumulative frequency (cf) just greater than $\dfrac{n}{2}(.$ i.e., $.\dfrac{100}{2}=50)$ is 78 , belonging to interval $35-40$.
Therefore, median class $=35-40$
Lower limit $(I)$ of median class $=35$
Class size $(h)=5$
Frequency $(f)$ of median class $=33$
Cumulative frequency ( $c f$ ) of class preceding median class $=45$
$ \begin{aligned} \text{ Median } & =l+(\dfrac{\dfrac{n}{2}-c f}{f}) \times h \\ & =35+(\dfrac{50-45}{33}) \times 5 \\ & =35+\dfrac{25}{33} \\ & =35.76 \end{aligned} $
Therefore, median age is 35.76 years.
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 :
| Length (in mm) | Number of leaves |
|---|---|
| $118-126$ | 3 |
| $127-135$ | 5 |
| $136-144$ | 9 |
| $145-153$ | 12 |
| $154-162$ | 5 |
| $163-171$ | 4 |
| $172-180$ | 2 |
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, …, 171.5 - 180.5.)
Show Answer
Solution
The given data does not have continuous class intervals. It can be observed that the difference between two class
intervals is 1. Therefore, $\dfrac{1}{2}=0.5$ has to be added and subtracted to upper class limits and lower class limits respectively.
Continuous class intervals with respective cumulative frequencies can be represented as follows.
| Length (in mm) | Number or leaves $\boldsymbol{{}f}_i$ | Cumulative frequency |
|---|
| $117.5-126.5$ | 3 | 3 |
|---|---|---|
| $126.5-135.5$ | 5 | $3+5=8$ |
| $135.5-144.5$ | 9 | $8+9=17$ |
| $144.5-153.5$ | 12 | $17+12=29$ |
| $153.5-162.5$ | 5 | $29+5=34$ |
| $162.5-171.5$ | 4 | $34+4=38$ |
| $171.5-180.5$ | 2 | $38+2=40$ |
From the table, it can be observed that the cumulative frequency just greater than $\dfrac{2}{2}(.$ i.e., $.\dfrac{40}{2}=20)$ is 29 , belonging to class interval 144.5 - 153.5 .
Median class $=144.5-153.5$
Lower limit $(I)$ of median class $=144.5$
Class size $(h)=9$
Frequency $(f)$ of median class $=12$
Cumulative frequency ( $c f$ ) of class preceding median class $=17$
Median
$ =l+(\dfrac{\dfrac{n}{2}-c f}{f}) \times h $
$=144.5+(\dfrac{20-17}{12}) \times 9$
$=144.5+\dfrac{9}{4}=146.75$
Therefore, median length of leaves is $146.75 mm$.
5. The following table gives the distribution of the life time of 400 neon lamps :
| Life time (in hours) | Number of lamps |
|---|---|
| $1500-2000$ | 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.
Show Answer
Solution
Thecumulative frequencies with their respective class intervals are as follows.
| Life time | Number of lamps $(\boldsymbol{{}f} _{\boldsymbol{{}i}})$ | Cumulative frequency |
|---|---|---|
| $1500-2000$ | 14 | 14 |
| $2000-2500$ | 56 | $14+56=70$ |
| $2500-3000$ | 60 | $70+60=130$ |
| $3000-3500$ | 86 | $130+86=216$ |
| $3500-4000$ | 74 | $216+74=290$ |
| $4000-4500$ | 62 | $290+62=352$ |
| $4500-5000$ | 48 | $352+48=400$ |
| Total $(n)$ | 400 |
It can be observed that the cumulative frequency just greater than
$ \dfrac{n}{2}(\text{ i.e., } \dfrac{400}{2}=200) $
interval 3000 - 3500.
Median class $=3000-3500$
Lower limit $(I)$ of median class $=3000$
Frequency $(f)$ of median class $=86$
Cumulative frequency ( $c$ ) of class preceding median class $=130$
Class size $(h)=500$
Median
$ =l+(\dfrac{\dfrac{n}{2}-c f}{f}) \times h $
$ \begin{aligned} & =3000+(\dfrac{200-130}{86}) \times 500 \\ & =3000+\dfrac{70 \times 500}{86} \end{aligned} $
$=3406.976$
Therefore, median life time of lamps is 3406.98 hours.
6. 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.
Show Answer
Solution
The cumulative frequencies with their respective class intervals are as follows.
| Number of letters | Frequency $(\boldsymbol{{}f} _{\boldsymbol{{}i}})$ | Cumulative frequency |
|---|---|---|
| $1-4$ | 6 | 6 |
| $4-7$ | 30 | $30+6=36$ |
| $7-10$ | 40 | $36+40=76$ |
| $10-13$ | 16 | $76+16=92$ |
| $13-16$ | 4 | $92+4=96$ |
| $16-19$ | 4 | $96+4=100$ |
| Total $(n)$ | 100 |
It can be observed that the cumulative frequency just greater than
$ \dfrac{n}{2}(\text{ i.e., } \dfrac{100}{2}=50) $
interval 7 - 10 .
Median class $=7-10$
Lower limit (I) of median class $=7$
Cumulative frequency ( $c f$ ) of class preceding median class $=36$
Frequency $(f)$ of median class $=40$
Class size $(h)=3$
Median
$ =l+(\dfrac{\dfrac{n}{2}-c f}{f}) \times h $
$=7+(\dfrac{50-36}{40}) \times 3$
$=7+\dfrac{14 \times 3}{40}$
$=8.05$
To find the class marks of the given class intervals, the following relation is used.
Class mark $=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Taking 11.5 as assumed mean (a), $d_i, u_i$, and $f_i u_i$ are calculated according to step deviation method as follows.
| Number of letters | Number of surnames $\boldsymbol{{}f}_i$ |
$\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{1 1 . 5}$ | $\boldsymbol{{}u}_i=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{3}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}u} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $1-4$ | 6 | 2.5 | -9 | -3 | -18 |
| $4-7$ | 30 | 5.5 | -6 | -2 | -60 |
| $7-10$ | 40 | 8.5 | -3 | -1 | -40 |
| $10-13$ | 16 |
7. 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 |
Show Answer
Solution
The cumulative frequencies with their respective class intervals are as follows.
| Weight (in kg) | Frequency (fi) | Cumulative frequency |
|---|---|---|
| $40-45$ | 2 | 2 |
| $45-50$ | 3 | $2+3=5$ |
| $50-55$ | 8 | $5+8=13$ |
|---|---|---|
| $55-60$ | 6 | $13+6=19$ |
| $60-65$ | 6 | $19+6=25$ |
| $65-70$ | 3 | $25+3=28$ |
| $70-75$ | 2 | $28+2=30$ |
| Total $(n)$ | 30 |
Cumulative frequency just greater than $\dfrac{n}{2}(.$ i.e., $.\dfrac{30}{2}=15)$ is 19 , belonging to class interval $55-60$.
Median class $=55-60$
Lower limit ( $/$ ) of median class $=55$
Frequency $(f)$ of median class $=6$
Cumulative frequency ( $c f$ ) of median class $=13$
Class size $(h)=5$
Median
$ =l+(\dfrac{\dfrac{n}{2}-c f}{f}) \times h $
$=55+(\dfrac{15-13}{6}) \times 5$
$=55+\dfrac{10}{6}$
$=56.67$
Therefore, median weight is $56.67 kg$.
