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Find 'mean' and 'mode' of the following data: Class: 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 Frequency: 11 | 8 | 15 | 7 | 10 | 9

Find 'mean' and 'mode' of the following data:
Class: 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 
Frequency: 11 | 8 | 15 | 7 | 10 | 9

This Question has 1 answers.

Class Intervals: $0-15, 15-30, 30-45, $
$45-60, 60-75, 75-90$

Frequencies: $11, 8, 15, 7, 10, 9$

\textbf{Step 1: Find the Mean}

Class midpoints $(x_i)$:

$ x_1 = 7.5, x_2 = 22.5, x_3 = 37.5, $
$x_4 = 52.5, x_5 = 67.5, x_6 = 82.5 $

Calculate $f_i x_i$:

$ (11 \times 7.5) + (8 \times 22.5) + (15 \times 37.5) + $
$(7 \times 52.5) + (10 \times 67.5) + (9 \times 82.5) $

$ = 82.5 + 180 + 562.5 + 367.5 + 675 + 742.5 = 2610 $

Total frequency:

$ \sum f_i = 11 + 8 + 15 + 7 + 10 + 9 = 60 $

Mean formula:

$ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} $

$ \bar{x} = \frac{2610}{60} = 43.5 $

\textbf{Step 2: Find the Mode}

The highest frequency is $15$, corresponding to the class $30-45$.

Using the mode formula:

$ \text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h $

Where:
$ L = 30 $,
$ f_1 = 15 $,
$ f_0 = 8 $,
$ f_2 = 7 $,
$ h = 15 $

Substituting the values:

$ \text{Mode} = 30 + \frac{(15 - 8)}{(2(15) - 8 - 7)} \times 15 $

$ = 30 + \frac{7}{30 - 15} \times 15 $

$ = 30 + \frac{7}{15} \times 15 $

$ = 30 + 7 $

$ = 37 $

Mean = $43.5$, Mode = $37$.

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