Chapter 13 Statistics Exercise-02
EXERCISE 13.2
1. 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.
Show Answer
Solution
To find the class marks $(x_i)$, the following relation is used.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Taking 30 as assumed mean (a), $d_i$ and $f_i d_i$ are calculated as follows.
| Age (in years) | Number of patients $\boldsymbol{{}f}_i$ | Class mark $\boldsymbol{{}x}_i$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{3 0}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}d} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|
| $5-15$ | 6 | 10 | -20 | -120 |
| $15-25$ | 11 | 20 | -10 | -110 |
| $25-35$ | 21 | 30 | 0 | 0 |
| $35-45$ | 23 | 40 | 10 | 230 |
| $45-55$ | 14 | 50 | 20 | 280 |
| $55-65$ | 5 | 60 | 30 | 150 |
| Total | 80 | 430 |
From the table, we obtain
$\sum f_i=80$
$\sum f_i d_i=430$
Mean, $\bar{{}x}=a+\dfrac{\sum f_i d_i}{\sum f_i}$
$ \begin{aligned} & =30+(\dfrac{430}{80}) \\ & =30+5.375 \\ & =35.375 \\ & \simeq 35.38 \end{aligned} $
Mean of this data is 35.38 . It represents that on an average, the age of a patient admitted to hospital was 35.38 years.
It can be observed that the maximum class frequency is 23 belonging to class interval 35 - 45 .
Modal class $=35-45$
Lower limit ( $l$ ) of modal class $=35$
Frequency $(f_1)$ of modal class $=23$
Class size $(h)=10$
Frequency $(f_0)$ of class preceding the modal class $=21$
Frequency $(f_2)$ of class succeeding the modal class $=14$
Mode $=$
$ l+(\dfrac{f_1-f_0}{2 f_1-f_0-f_2}) \times h $
$=35+(\dfrac{23-21}{2(23)-21-14}) \times 10$
$=35+[\dfrac{2}{46-35}] \times 10$
$=35+\dfrac{20}{11}$
$=35+1.81$
$=36.8$
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
| Lifetimes (in hours) | $0-20$ | $20-40$ | $40-60$ | $60-80$ | $80-100$ | $100-120$ |
|---|---|---|---|---|---|---|
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Show Answer
Solution
From the data given above, it can be observed that the maximum class frequency is 61 , belonging to class interval 60 - 80.
Therefore, modal class $=60-80$
Lower class limit $(l)$ of modal class $=60$
Frequency $(f_1)$ of modal class $=61$
Frequency $(f_0)$ of class preceding the modal class $=52$
Frequency $(f_2)$ of class succeeding the modal class $=38$
Class size $(h)=20$
Mode $=l+(\dfrac{f_1-f_0}{2 f_1-f_0-f_2}) \times h$
$=60+(\dfrac{61-52}{2(61)-52-38})(20)$
$ \begin{aligned} & =60+(\dfrac{9}{122-90})(20) \\ & =60+(\dfrac{9 \times 20}{32}) \\ & =60+\dfrac{90}{16}=60+5.625 \\ & =65.625 \end{aligned} $
Therefore, modal lifetime of electrical components is 65.625 hours.
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 :
| Expenditure (in ₹) | Number of families |
|---|---|
| $1000-1500$ | 24 |
| $1500-2000$ | 40 |
| $2000-2500$ | 33 |
| $2500-3000$ | 28 |
| $3000-3500$ | 30 |
| $3500-4000$ | 22 |
| $4000-4500$ | 16 |
| $4500-5000$ | 7 |
Show Answer
Solution
It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 - 2000 intervals.
Therefore, modal class $=1500-2000$
Lower limit $(l)$ of modal class $=1500$
Frequency $(f_1)$ of modal class $=40$
Frequency $(f_0)$ of class preceding modal class $=24$
Frequency $(f_2)$ of class succeeding modal class $=33$
Class size $(h)=500$
$ \begin{aligned} \text{ Mode } & =l+(\dfrac{f_1-f_0}{2 f_1-f_0-f_2}) \times h \\ & =1500+(\dfrac{40-24}{2(40)-24-33}) \times 500 \\ & =1500+(\dfrac{16}{80-57}) \times 500 \\ & =1500+\dfrac{8000}{23} \\ & =1500+347.826 \\ & =1847.826=1847.83 \end{aligned} $
Therefore, modal monthly expenditure was Rs 1847.83.
To find the class mark, the following relation is used.
Class mark $=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Class size $(h)$ of the given data $=500$
Taking 2750 as assumed mean (a), $d_i, u_i$, and $f_i u_i$ are calculated as follows.
| Expenditure (in Rs.) | No. of families | Class Mark $\left(x_i\right)$ | $d_i=x_i-a$ | $u_i=\frac{d_i}{500}$ | $f_i u_i$ |
|---|---|---|---|---|---|
| $1000-1500$ | 24 | 1250 | -2000 | -4 | -96 |
| $1500-2000$ | 40 | 1750 | -1500 | -3 | -120 |
| $2000-2500$ | 33 | 2250 | -1000 | -2 | -66 |
| $2500-3000$ | 28 | 2750 | -500 | -1 | -28 |
| $3000-3500$ | 30 | $3250=a$ | 0 | 0 | 0 |
| $3500-4000$ | 22 | 3750 | 500 | 1 | 22 |
| $4000-4500$ | 16 | 4250 | 1000 | 2 | 32 |
| $4500-5000$ | 7 | 4750 | 1500 | 3 | 21 |
| Total | $\Sigma f_i= 200$ | $\Sigma f_i u_i=-235$ |
From the table, we obtatain
$\begin{aligned} & \sum f_i=200, \sum u_i f_i=-235 \\ & \text { Mean monthly income }=\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma f_i u_i}{\Sigma f_i} \times \mathrm{h} \\ & =3250-\frac{235}{200} \times 500 \\ & =3250-587.5 \\ & =\text { Rs. } 2662.50\end{aligned}$
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.
| 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 |
Show Answer
Solution
It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 - 35 .
Therefore, modal class $=30-35$
Class size $(h)=5$
Lower limit $(l)$ of modal class $=30$
Frequency $(f_1)$ of modal class $=10$
Frequency $(f_0)$ of class preceding modal class $=9$
Frequency $(f_2.$ ) of class succeeding modal class $=3$
$ \begin{aligned} \text{ Mode } & =l+(\dfrac{f_1-f_0}{2 f_1-f_0-f_2}) \times h \\ & =30+(\dfrac{10-9}{2(10)-9-3}) \times(5) \\ & =30+(\dfrac{1}{20-12}) 5 \\ & =30+\dfrac{5}{8}=30.625 \end{aligned} $
Mode $=30.6$
It represents that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
Class mark $=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Taking 32.5 as assumed mean (a), $d_i, u_i$, and $f_i u_i$ are calculated as follows.
| Number of students per teacher | Number of states/U.T $(f_i)$ |
$\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d}_i=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{3 2 . 5}$ | $\boldsymbol{{}u} _{\boldsymbol{{}i}}=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{5}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} u _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $15-20$ | 3 | 17.5 | -15 | -3 | -9 |
| $20-25$ | 8 | 22.5 | -10 | -2 | -16 |
| $25-30$ | 9 | 27.5 | -5 | -1 | -9 |
| $30-35$ | 10 | 32.5 | 0 | 0 | 0 |
| $35-40$ | 3 | 37.5 | 5 | 1 | 3 |
| $40-45$ | 0 | 42.5 | 10 | 2 | 0 |
| $45-50$ | 0 | 47.5 | 15 | 3 | 0 |
| $50-55$ | 2 | 52.5 | 20 | 4 | 8 |
| Total | 35 | -23 |
Mean, $\bar{{}x}=a+(\dfrac{\sum f_i u_i}{\sum f_i}) h$
$ \begin{aligned} & =32.5+(\dfrac{-23}{35}) \times 5 \\ & =32.5-\dfrac{23}{7}=32.5-3.28 \\ & =29.22 \end{aligned} $
Therefore, mean of the data is 29.2.
It represents that
- The maximum number of students per teacher is 30.6
- The average number of students per teacher is 29.2
5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
| Runs scored | Number of batsmen |
|---|---|
| $3000-4000$ | 4 |
| $4000-5000$ | 18 |
| $5000-6000$ | 9 |
| $6000-7000$ | 7 |
| $7000-8000$ | 6 |
| $8000-9000$ | 3 |
| $9000-10000$ | 1 |
| $10000-11000$ | 1 |
Find the mode of the data.
Show Answer
Solution
From the given data, it can be observed that the maximum class frequency is 18 , belonging to class interval $4000-5000$ .
Therefore, modal class $=4000-5000$
Lower limit (I) of modal class $=4000$
Frequency $(f_1)$ of modal class $=18$
Frequency $(f_0)$ of class preceding modal class $=4$
Frequency $(f_2.$ ) of class succeeding modal class $=9$
Class size $(h)=1000$
Mode $=l+(\dfrac{f_1-f_0}{2 f_1-f_0-f_2}) \times h$
$=4000+(\dfrac{18-4}{2(18)-4-9}) \times 1000$
$=4000+(\dfrac{14000}{23})$
$=4000+608.695$
$=4608.695$
Therefore, mode of the given data is 4608.7 runs.
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 :
| Number of cars | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
|---|---|---|---|---|---|---|---|---|
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Show Answer
Solution
From the given data, it can be observed that the maximum class frequency is 20 , belonging to 40 - 50 class intervals.
Therefore, modal class $=40-50$
Lower limit ( $l$ ) of modal class $=40$
Frequency $(f_1)$ of modal class $=20$
Frequency $(f_0)$ of class preceding modal class $=12$
Frequency $(f_2)$ of class succeeding modal class $=11$
Class size $=10$
$ \begin{aligned} \text{ Mode } & =l+(\dfrac{f_1-f_0}{2 f_1-f_0-f_2}) \times h \\ & =40+[\dfrac{20-12}{2(20)-12-11}] \times 10 \\ & =40+(\dfrac{80}{40-23}) \\ & =40+\dfrac{80}{17} \end{aligned} $
$=40+4.7$
$=44.7$
Therefore, mode of this data is 44.7 cars.
