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:
$\begin{array}{|c|c|} \hline \text{Expenditure (in Rs)} & \text{Number of families} \\ \hline 1000-1500 & 24 \\ 1500-2000 & 40 \\ 2000-2500 & 33 \\ 2500-3000 & 28 \\ 3000-3500 & 30 \\ 3500-4000 & 22 \\ 4000-4500 & 16 \\ 4500-5000 & 7 \\ \hline \end{array}$
$\begin{array}{|c|c|} \hline \text{Expenditure (in Rs)} & \text{Number of families} \\ \hline 1000-1500 & 24 \\ 1500-2000 & 40 \\ 2000-2500 & 33 \\ 2500-3000 & 28 \\ 3000-3500 & 30 \\ 3500-4000 & 22 \\ 4000-4500 & 16 \\ 4500-5000 & 7 \\ \hline \end{array}$
(N/A) It can be observed from the given data that the maximum class frequency is $40$,belonging to $1500-2000$ interval.
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 $(h) = 500$.
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$= 1500 + \left( \frac{40 - 24}{2(40) - 24 - 33} \right) \times 500$
$= 1500 + \left( \frac{16}{80 - 57} \right) \times 500 = 1500 + \frac{8000}{23} \approx 1500 + 347.83 = 1847.83$.
Thus,the modal monthly expenditure is Rs. $1847.83$.
To find the mean,we use the step-deviation method:
Mean $(\bar{x}) = a + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right)$
Using $a = 2750$ and $h = 500$,we calculate $\sum f_i u_i = -35$ and $\sum f_i = 200$.
Mean $= 2750 + 500 \times \left( \frac{-35}{200} \right) = 2750 - 87.5 = 2662.5$.
Thus,the mean monthly expenditure is Rs. $2662.5$.
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 $(h) = 500$.
Mode $= l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$= 1500 + \left( \frac{40 - 24}{2(40) - 24 - 33} \right) \times 500$
$= 1500 + \left( \frac{16}{80 - 57} \right) \times 500 = 1500 + \frac{8000}{23} \approx 1500 + 347.83 = 1847.83$.
Thus,the modal monthly expenditure is Rs. $1847.83$.
To find the mean,we use the step-deviation method:
Mean $(\bar{x}) = a + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right)$
Using $a = 2750$ and $h = 500$,we calculate $\sum f_i u_i = -35$ and $\sum f_i = 200$.
Mean $= 2750 + 500 \times \left( \frac{-35}{200} \right) = 2750 - 87.5 = 2662.5$.
Thus,the mean monthly expenditure is Rs. $2662.5$.
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Find the mode of the data.
| 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.
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View SolutionThe median of the following data is $525$. Find the values of $x$ and $y$,if the total frequency is $100$.
Class interval Frequency $0-100$ $2$ $100-200$ $5$ $200-300$ $x$ $300-400$ $12$ $400-500$ $17$ $500-600$ $20$ $600-700$ $y$ $700-800$ $9$ $800-900$ $7$ $900-1000$ $4$
| Class interval | Frequency |
| $0-100$ | $2$ |
| $100-200$ | $5$ |
| $200-300$ | $x$ |
| $300-400$ | $12$ |
| $400-500$ | $17$ |
| $500-600$ | $20$ |
| $600-700$ | $y$ |
| $700-800$ | $9$ |
| $800-900$ | $7$ |
| $900-1000$ | $4$ |
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Find the median length of the leaves.
| 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.
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Find the mean daily wages of the workers of the factory by using an appropriate method.
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Find the mean daily wages of the workers of the factory by using an appropriate method.
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Which method did you use for finding the mean,and why?
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| Number of houses | $1$ | $2$ | $1$ | $5$ | $6$ | $2$ | $3$ |
Which method did you use for finding the mean,and why?
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