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The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food.

Daily Expenditure (in Rs.) 100-150 150-200 200-250 250-300 300-350
Number of Households 4 5 12 2 2

 

 

 

 

 
 
 
 
 

Answers (1)

 

Daily Expenditure No of households (fi) Class Mark (xi) xi - 225 \mathrm{u_i = \frac{x_i-2250}{50}} fiui
100-150 4 \mathrm{\frac{100+150}{2} = 125} 125-220=-100 \mathrm{\frac{-100}{50}=-2} \mathrm{4\times (-2) = -8}
150-200 5 \mathrm{\frac{150+200 = 175}{2}} 175-225=-50 \mathrm{\frac{-50}{50}=-1} \mathrm{5\times (-1) = -5}
200-250 12 \mathrm{\frac{200+250}{2} = 225} 225-225=0 \mathrm{\frac{0}{50}=0} \mathrm{12\times (0) = 0}
250-300 2 \mathrm{\frac{250+300}{2} = 275} 275-225=50 \mathrm{\frac{50}{50}=1} \mathrm{2\times 1 = 2}
300-350 2 \mathrm{\frac{300+350}{2}=325} 325-225=100 \mathrm{\frac{100}{50}=2} \mathrm{2\times 2 = 4}
  \mathrm{\Sigma f_i = 25}       \mathrm{\Sigma f_ix_i = -7}

As we can see the highest frequency (fi) = 12

The class for (fi = 12) \Rightarrow 200-250

Now,    \mathrm{\text{Mean }\bar{x} = a +h\times\frac{\Sigma f_i u_i}{\Sigma f_i}}

            Here a = assumed mean = \mathrm{\frac{200 + 250}{2} = 225}

                     h = class interval = 150-100 = 50

                     \mathrm{u_i = \frac{x_i - a}{h}}

                \mathrm{\text{Mean }\bar{x} = 225 +50\times \frac{-7}{25}}

                    \mathrm{\Rightarrow \bar{x} = 225 +2\times (-7) }

                                \mathrm{= 225 -14 }

                     \mathrm{\Rightarrow \bar{x}= 211 }

    Mean Daily expenditure = 211 Rupees.

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Safeer PP

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