UPSKILL MATH PLUS
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Learn moreThe mean of the ungrouped frequency distribution can be determined using the formula:
\(\overline X = \frac{f_1 x_1 + f_2 x_2 + ... + f_n x_n}{f_1 + f_2 + ... + f_n}\) \(= \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}\)
Example:
The height(in \(cm\)) of \(20\) students in a classroom are:
Height \(x_i\) | Number of students \(f_i\) |
| \(130\) | \(1\) |
| \(135\) | \(2\) |
| \(140\) | \(1\) |
| \(155\) | \(2\) |
| \(163\) | \(1\) |
| \(165\) | \(3\) |
| \(177\) | \(2\) |
| \(189\) | \(2\) |
| \(196\) | \(2\) |
| \(100\) | \(4\) |
Find the mean height of the \(20\) students.
Solution:
To find the value of \(f_ix_i\), multiply the value of \(x\) and \(f\) of each entry.
Consider for the mark \(130\). That is, \(130 \times 1 = 130\)
Similarly, for the mark \(135\), we have \(135 \times 2 = 270\) and so on.
Tabulating these values, we get:
Marks \(x_i\) | Frequency \(f_i\) | \(f_ix_i\) |
| \(130\) | \(1\) | \(130\) |
| \(135\) | \(2\) | \(270\) |
| \(140\) | \(1\) | \(140\) |
| \(155\) | \(2\) | \(310\) |
| \(163\) | \(1\) | \(163\) |
| \(165\) | \(3\) | \(495\) |
| \(177\) | \(2\) | \(354\) |
| \(189\) | \(2\) | \(378\) |
| \(196\) | \(2\) | \(392\) |
| \(100\) | \(4\) | \(400\) |
| Total | \(\sum f_i = 20\) | \(\sum f_ix_i = 3032\) |
Substituting the known values in the above formula, we get:
Mean \(\overline X = \frac{3032}{20}\) \(= 151.6\)
Therefore, the mean of the given data is \(151.6\).
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