Consider the following distribution.
|
Class Interval |
0-20 |
20-40 |
40-60 |
60-80 |
80-100 |
|
Frequency |
17 |
28 |
32 |
f |
19 |
f the mean of the above distribution is 50, find the value of f.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 24
To find the value of $f$, we use the formula for the mean of a frequency distribution:
$\text{Mean } (\bar{x}) = \frac{\sum f_i x_i}{\sum f_i}$
where $f_i$ is the frequency and $x_i$ is the class mark (midpoint) of each interval.
Step 1: Identify Class Marks ($x_i$) and $f_i x_i$
Assuming the distribution follows a uniform pattern of class width 20 (as seen in the later intervals), the first interval is likely 0-20. Let's calculate the midpoints based on this pattern:
|
Class Interval |
Frequency ($f_i$) |
Class Mark ($x_i$) |
$f_ix_i$ |
|
0 - 20 |
17 |
10 |
$17 \times 10 = 170$ |
|
20 - 40 |
28 |
30 |
$28 \times 30 = 840$ |
|
40 - 60 |
32 |
50 |
$32 \times 50 = 1600$ |
|
60 - 80 |
$f$ |
70 |
$f \times 70 = 70f$ |
|
80 - 100 |
19 |
90 |
$19 \times 90 = 1710$ |
|
Total |
$\sum f_i = 96 + f$ |
$\sum f_i x_i = 4320 + 70f$ |
Step 2: Use the Mean to find $f$
The given mean is 50.
$50 = \frac{4320 + 70f}{96 + f}$
Multiply both sides by $(96 + f)$:
$50(96 + f) = 4320 + 70f$
$4800 + 50f = 4320 + 70f$
Rearrange the terms to solve for $f$:
$4800 - 4320 = 70f - 50f$
$480 = 20f$
$f = \frac{480}{20} = 24$