Chapter 13 Statistics Exercise-01
EXERCISE 13.1
1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
| Number of plants | $0-2$ | $2-4$ | $4-6$ | $6-8$ | $8-10$ | $10-12$ | $12-14$ |
|---|---|---|---|---|---|---|---|
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Which method did you use for finding the mean, and why?
Show Answer
Solution
To find the class mark $(x_i)$ for each interval, the following relation is used.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
$x_i$ andf $f_i$ can be calculated as follows.
| Number of plants | Number of houses $(\boldsymbol{{}f} _{\boldsymbol{{}i})}.$ |
$\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}x}_i$ |
|---|---|---|---|
| $0-2$ | 1 | 1 | $1 \times 1=1$ |
| $2-4$ | 2 | 3 | $2 \times 3=6$ |
| $4-6$ | 1 | 5 | $1 \times 5=5$ |
| $6-8$ | 5 | 7 | $5 \times 7=35$ |
| $8-10$ | 6 | 9 | $6 \times 9=54$ |
| $10-12$ | 2 | 11 | $2 \times 11=22$ |
| $12-14$ | 3 | $3 \times 13=39$ | |
| Total | 20 | 162 |
From the table, it can be observed that $\sum f_i=20$
$\sum f_i x_i=162$
Mean,
$\bar{{}x}=\dfrac{\sum f_i x_i}{\sum f_i}$
$=\dfrac{162}{20}=8.1$
Therefore, mean number of plants per house is 8.1.
Here, direct method has been used as the values of class marks $(x_i)$ and $f_i$ are small.
2. Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages (in ₹) | $500-520$ | $520-540$ | $540-560$ | $560-580$ | $580-600$ |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Show Answer
Solution
To find the class mark for each interval, the following relation is used.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Class size $(h)$ of this data $=20$
Taking 150 as assured mean (a), $d_i, u_i$, and $f_i u_i$ can be calculated as follows.
| Daily wages (in Rs) |
Number of workers $(f_i)$ | $\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{1 5 0}$ | $\boldsymbol{{}u} _{\boldsymbol{{}i}}=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{2 0}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $100-120$ | 12 | 110 | -40 | -2 | -24 |
| $120-140$ | 14 | 130 | -20 | -1 | -14 |
| $140-160$ | 8 | 150 | 0 | 0 | 0 |
| $160-180$ | 6 | 170 | 20 | 1 | 6 |
| $180-200$ | 10 | 190 | 40 | 2 | 20 |
| Total | 50 | -12 |
From the table, it can be observed that
$ \begin{aligned} \sum f_i & =50 \\ \sum f_i u_i & =-12 \\ \text{ Mean } \bar{{}x} & =a+(\dfrac{\sum f_i u_i}{\sum f_i}) h \\ & =150+(\dfrac{-12}{50}) 20 \\ & =150-\dfrac{24}{5} \\ & =150-4.8 \\ & =145.2 \end{aligned} $
Therefore, the mean daily wage of the workers of the factory is Rs 145.20.
3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency $f$.
| Daily pocket allowance (in ₹) | $11-13$ | $13-15$ | $15-17$ | $17-19$ | $19-21$ | $21-23$ | $23-25$ |
|---|---|---|---|---|---|---|---|
| Number of children | 7 | 6 | 9 | 13 | $f$ | 5 | 4 |
Show Answer
Solution
To find the class mark $(x_i)$ for each interval, the following relation is used.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Given that, mean pocket allowance, $\bar{{}x}=$ Rs 18
Taking 18 as assured mean (a), $d_i$ and $f_i d_i$ are calculated as follows.
| Daily pocket allowance (in Rs) |
Number of children $\boldsymbol{{}f} _{\boldsymbol{{}i}}$ |
Class mark $\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{1 8}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|
| $11-13$ | 7 | 12 | -6 | -42 |
| $13-15$ | 6 | 14 | -4 | -24 |
| $15-17$ | 9 | 16 | -2 | -18 |
| $17-19$ | 13 | 18 | 0 | 0 |
| $19-21$ | $f$ | 20 | 2 | $2 f$ |
| $21-23$ | 5 | 22 | 4 | 20 |
| $23-25$ | 4 | 24 | 6 | 24 |
| Total | $\sum f_i=44+f$ | $2 f-40$ |
From the table, we obtain
$\sum f_i=44+f$
$\sum f_i u_i=2 f-40$
$\bar{{}x}=a+\dfrac{\sum f_i d_i}{\sum f_i}$
$18=18+(\dfrac{2 f-40}{44+f})$
$0=(\dfrac{2 f-40}{44+f})$
$2 f-40=0$
$2 f=40$
$f=20$
Hence, the missing frequency, $f$, is 20 .
4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
| Number of heartbeats per minute | $65-68$ | $68-71$ | $71-74$ | $74-77$ | $77-80$ | $80-83$ | $83-86$ |
|---|---|---|---|---|---|---|---|
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Show Answer
Solution
To find the class mark of each interval $(x_i)$, the following relation is used.
$ x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2} $
Class size, $h$, of this data $=3$
Taking 75.5 as assumed mean (a), di, $u_i, f_i u_i$ are calculated as follows.
| Number of heart beats per minute | Number of women $\boldsymbol{{}f}_i$ |
$\boldsymbol{{}x}_i$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{7 5 . 5}$ | $\boldsymbol{{}u} _{\boldsymbol{{}i}}=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{3}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}u} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $65-68$ | 2 | 66.5 | -9 | -3 | -6 |
| $68-71$ | 4 | 69.5 | -6 | -2 | -8 |
| $71-74$ | 3 | 72.5 | -3 | -1 | -3 |
| $74-77$ | 8 | 75.5 | 0 | 0 | 0 |
| $77-80$ | 7 | 78.5 | 3 | 1 | 7 |
| $80-83$ | 4 | 81.5 | 6 | 2 | 8 |
| $83-86$ | 2 | 84.5 | 9 | 3 | 6 |
| Total | 30 | 4 |
From the table, we obtain
$\sum f_i=30$
$\sum f_i u_i=4$
Mean $\bar{{}x}=a+(\dfrac{\sum f_i u_i}{\sum f_i}) \times h$
$ \begin{aligned} & =75.5+(\dfrac{4}{30}) \times 3 \\ & =75.5+0.4=75.9 \end{aligned} $
Therefore, mean hear beats per minute for these women are 75.9 beats per minute.
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
| Number of mangoes | $50-52$ | $53-55$ | $56-58$ | $59-61$ | $62-64$ |
|---|---|---|---|---|---|
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Show Answer
Solution
| $50-52$ | 15 |
|---|---|
| $53-55$ | 110 |
| $56-58$ | 135 |
| $59-61$ | 115 |
| $62-64$ | 25 |
It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore,
$\dfrac{1}{2}$ has to be added to the upper class limit and $\dfrac{1}{2}$ has to be subtracted from the lower class limit of each interval.
Class mark $(x_i)$ can be obtained by using the following relation.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Class size $(h)$ of this data $=3$
Taking 57 as assumed mean (a), $d_i, u_i, f_i u_i$ are calculated as follows.
| Class interval | $\boldsymbol{{}f}_i$ | $\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{5 7}$ | $\boldsymbol{{}u} _{\boldsymbol{{}i}}=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{3}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}u} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $49.5-52.5$ | 15 | 51 | -6 | -2 | -30 |
| $52.5-55.5$ | 110 | 54 | -3 | -1 | -110 |
| $55.5-58.5$ | 135 | 57 | 0 | 0 | 0 |
| $58.5-61.5$ | 115 | 60 | 3 | 1 | 115 |
| $61.5-64.5$ | 25 | 63 | 6 | 2 | 50 |
| Total | 400 | 25 |
It can be observed that
$\sum f_i=400$
$\sum f_i u_i=25$
$ \begin{aligned} \text{ Mean, } \bar{{}x} & =a+(\dfrac{\sum f_i u_i}{\sum f_i}) \times h \\ & =57+(\dfrac{25}{400}) \times 3 \\ & =57+\dfrac{3}{16}=57+0.1875 \\ & =57.1875 \\ & \simeq 57.19 \end{aligned} $
Mean number of mangoes kept in a packing box is 57.19.
Step deviation method is used here as the values of $f_i, d_i$ are big and also, there is a common factor of all $d_i$ ’s
6. The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure (in ₹) | $100-150$ | $150-200$ | $200-250$ | $250-300$ | $300-350$ |
|---|---|---|---|---|---|
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
Show Answer
Solution
To find the class mark $(x_i)$ for each interval, the following relation is used.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Class size $=50$
Taking 225 as assumed mean (a), $d_i, u_i, f_i u_i$ are calculated as follows.
| Daily expenditure (in Rs) | $\boldsymbol{{}f} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{2 2 5}$ | $\boldsymbol{{}u} _{\boldsymbol{{}i}}=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{5 0}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}u} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $100-150$ | 4 | 125 | -100 | -2 | -8 |
| $150-200$ | 5 | 175 | -50 | -1 | -5 |
| $200-250$ | 12 | 225 | 0 | 0 | 0 |
| $250-300$ | 2 | 275 | 50 | 1 | 2 |
| $300-350$ | 2 | 325 | 100 | 2 | 4 |
| Total | 25 | -7 |
From the table, we obtain
$\sum f_i=25$
$\sum f_i u_i=-7$
$ \text{ Mean, } \begin{aligned} \bar{{}x} & =a+(\dfrac{\sum f_i u_i}{\sum f_i}) \times h \\ & =225+(\dfrac{-7}{25}) \times(50) \\ & =225-14 \\ & =211 \end{aligned} $
Therefore, mean daily expenditure on food is Rs 211 .
7. To find out the concentration of $\mathrm{SO}_{2}$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
| Concentration of $\mathrm{SO}_{2}$ (in ppm) | Frequency |
|---|---|
| $0.00-0.04$ | 4 |
| $0.04-0.08$ | 9 |
| $0.08-0.12$ | 9 |
| $0.12-0.16$ | 2 |
| $0.16-0.20$ | 4 |
| $0.20-0.24$ | 2 |
Find the mean concentration of $\mathrm{SO}_{2}$ in the air.
Show Answer
Solution
To find the class marks for each interval, the following relation is used.
$ x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2} $
Class size of this data $=0.04$
Taking 0.14 as assumed mean (a), $d_i, u_i, f_i u_i$ are calculated as follows.
| Concentration of $\mathbf{S O}_2$ (in ppm) | Frequency $\boldsymbol{{}f}_i$ | Class mark $\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{0 . 1 4}$ | $\boldsymbol{{}u} _{\boldsymbol{{}i}}=\dfrac{\boldsymbol{{}d} _{\boldsymbol{{}i}}}{\mathbf{0 . 0 4}}$ | $\boldsymbol{{}f} _{\boldsymbol{{}i}} \boldsymbol{{}u} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|---|
| $0.00-0.04$ | 4 | 0.02 | -0.12 | -3 | -12 |
| $0.04-0.08$ | 9 | 0.06 | -0.08 | -2 | -18 |
| $0.08-0.12$ | 9 | 0.10 | -0.04 | -1 | -9 |
| $0.12-0.16$ | 2 | 0.14 | 0 | 0 | 0 |
| $0.16-0.20$ | 4 | 0.18 | 0.04 | 1 | 4 |
| $0.20-0.24$ | 2 | 0.22 | 0.08 | 2 | 4 |
| Total | 30 | -31 |
From the table, we obtain
$\sum f_i=30$
$\sum f_i u_i=-31$
Mean, $\bar{{}x}=a+(\dfrac{\sum f_i u_i}{\sum f_i}) \times h$
$ \begin{aligned} & =0.14+(\dfrac{-31}{30})(0.04) \\ & =0.14-0.04133 \\ & =0.09867 \\ & \simeq 0.099 ppm \end{aligned} $
Therefore, mean concentration of $SO_2$ in the air is $0.099 ppm$.
8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
| Number of days | $0-6$ | $6-10$ | $10-14$ | $14-20$ | $20-28$ | $28-38$ | $38-40$ |
|---|---|---|---|---|---|---|---|
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Show Answer
Solution
To find the class mark of each interval, the following relation is used.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Taking 17 as assumed mean (a), $d_i$ and $f_i d_i$ are calculated as follows.
| Number of days | Number of students $\boldsymbol{{}f}_i$ | $\boldsymbol{{}x} _{\boldsymbol{{}i}}$ | $\boldsymbol{{}d} _{\boldsymbol{{}i}}=\boldsymbol{{}x} _{\boldsymbol{{}i}}-\mathbf{1 7}$ | $\boldsymbol{{}f i} _{\boldsymbol{{}i}}$ |
|---|---|---|---|---|
| $0-6$ | 11 | 3 | -14 | -154 |
| $6-10$ | 10 | 8 | -9 | -90 |
| $10-14$ | 7 | 12 | -5 | -35 |
| $14-20$ | 4 | 17 | 0 | 0 |
| $20-28$ | 4 | 24 | 7 | 28 |
| $28-38$ | 3 | 33 | 16 | 48 |
| $38-40$ | 1 | 39 | 22 | 22 |
| Total | 40 | -181 |
From the table, we obtain
$\sum f_i=40$
$\sum f_i d_i=-181$
Mean, $\bar{{}x}=a+(\dfrac{\sum f_i d_i}{\sum f_i})$
$=17+(\dfrac{-181}{40})$
$=17-4.525$
$=12.475$
$\simeq 12.48$
Therefore, the mean number of days is 12.48 days for which a student was absent.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | $45-55$ | $55-65$ | $65-75$ | $75-85$ | $85-95$ |
|---|---|---|---|---|---|
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Show Answer
Solution
To find the class marks, the following relation is used.
$x_i=\dfrac{\text{ Upper class limit }+ \text{ Lower class limit }}{2}$
Class size $(h)$ for this data $=10$
Taking 70 as assumed mean (a), $d_i, u_i$, and $f_i u_i$ are calculated as follows.
| Literacy rate (in %) | Number of cities | $x_i$ | $d_i=x_i-70$ | $u_i=\dfrac{d_i}{10}$ | $f_i u_i$ |
|---|---|---|---|---|---|
| $45-55$ | 3 | 50 | -20 | -2 | -6 |
| $55-65$ | 10 | 60 | -10 | -1 | -10 |
| $65-75$ | 11 | 70 | 0 | 0 | 0 |
| $75-85$ | 8 | 80 | 10 | 1 | 8 |
| $85-95$ | 3 | 90 | 20 | 2 | 6 |
| Total | 35 | -2 |
From the table, we obtain
$\sum f_i=35$
$\sum f_i u_i=-2$
$ \text{ Mean, } \begin{aligned} \bar{{}x} & =a+(\dfrac{\sum f_i u_i}{\sum f_i}) \times h \\ & =70+(\dfrac{-2}{35}) \times(10) \\ & =70-\dfrac{20}{35} \\ & =70-\dfrac{4}{7} \\ & =70-0.57 \\ & =69.43 \end{aligned} $
Therefore, mean literacy rate is $69.43 %$.
