i) Draw percentage bar diagram from the following data. [5] Item Food Clothing Rent Fuel Education Miscellaneous : : : : : : : Expenditure (Rs.) 4500 3500 3000 2500 4500 5000 ii) Calculate the…

i) Percentage Bar Diagram

Step 1: Calculate Total Expenditure

First, we sum up all the expenditure values to find the total expenditure.

$$4500+3500+3000+2500+4500+5000=23000\text{ Rs.}$$

Step 2: Calculate Percentage for Each Item

Next, we calculate the percentage of each item relative to the total expenditure.

$$\text{Percentage}=\left(\frac{\text{Expenditure}}{\text{Total Expenditure}}\right)\times 100$$

• Food: $\left(\frac{4500}{23000}\right)\times 100=19.57\%$
• Clothing: $\left(\frac{3500}{23000}\right)\times 100=15.22\%$
• Rent: $\left(\frac{3000}{23000}\right)\times 100=13.04\%$
• Fuel: $\left(\frac{2500}{23000}\right)\times 100=10.87\%$
• Education: $\left(\frac{4500}{23000}\right)\times 100=19.57\%$
• Miscellaneous: $\left(\frac{5000}{23000}\right)\times 100=21.74\%$

Step 3: Draw the Percentage Bar Diagram

Below is the percentage bar diagram representing the expenditure on different items.

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ii) Calculation of Median

Step 1: Find the Total Number of Students

The total number of students is given by the cumulative frequency of the last class interval.

$$90+85+68+48+33+26+20+15+12+4=391$$

Step 2: Determine the Median Position

The median is the value of the middle item in an ordered series. The position of the median is given by:

$$\text{Median Position}=\frac{N+1}{2}=\frac{391+1}{2}=196$$

Step 3: Find the Median Class

We now find the class interval in which the 196th item lies. We calculate the cumulative frequency to identify the correct class.

The cumulative frequency reaches 243 in the class interval 80-90, which includes the 196th item. Thus, the median class is 80-90.

Step 4: Apply the Median Formula

The formula for the median in a continuous series is:

$$\text{Median}=L+\left(\frac{\frac{N}{2}-C}{F}\right)\times W$$

Where:

• $L$ = Lower limit of the median class = 80
• $N$ = Total number of students = 391
• $C$ = Cumulative frequency of the class preceding the median class = 175
• $F$ = Frequency of the median class = 68
• $W$ = Width of the median class = 10 (since 90 - 80 = 10)

Substituting the values:

$$\text{Median}=80+\left(\frac{\frac{391}{2}-175}{68}\right)\times 10$$

$$=80+\left(\frac{195.5-175}{68}\right)\times 10$$

$$=80+\left(\frac{20.5}{68}\right)\times 10$$

$$=80+0.3\times 10$$

$$=80+3=83$$

Median = 83