Calculate the average (mean) from the following frequency distribution using any method: <table> <tr> <th>Class Interval</th> <td>85–105</td> <td>105–125</td> <td>125–145</td> <td>145–165</td> <td>165–185</td> <td>185–205</td> </tr> <tr> <th>Frequency</th> <td>3</td> <td>12</td> <td>18</td> <td>10</td> <td>5</td> <td>2</td> </tr> <tr> <th>Class Midpoint</th> <td>95</td> <td>115</td> <td>135</td> <td>155</td> <td>175</td> <td>195</td> </tr> <tr> <th>\(f_i x_i\)</th> <td>285</td> <td>1380</td> <td>2430</td> <td>1550</td> <td>875</td> <td>390</td> </tr> <tr> <th>Total</th> <td>\( \sum f_i = 50 \)</td> <td colspan="2"></td> <td colspan="3">\( \sum f_i x_i = 6910 \)</td> </tr> </table> > Mean using direct method = \(\frac{\sum f_i x_i}{\sum f_i} = \frac{6910}{50} = 138.2\)

Q. Calculate the average (mean) from the following frequency distribution using any method:

Class Interval	85–105	105–125	125–145	145–165	165–185	185–205
Frequency	3	12	18	10	5	2
Class Midpoint	95	115	135	155	175	195
\(f_i x_i\)	285	1380	2430	1550	875	390
Total	\( \sum f_i = 50 \)			\( \sum f_i x_i = 6910 \)

> Mean using direct method = \(\frac{\sum f_i x_i}{\sum f_i} = \frac{6910}{50} = 138.2\)

Class Interval	Frequency	Cumulative Frequency (Less than type)
0–10	8	8
10–20	7	15
20–30	14	29
30–40	13	42
40–50	18	60
50–60	10	70

Here, \(n = 70\), so \(\frac{n}{2} = \frac{70}{2} = 35\) The cumulative frequency just greater than 35 lies in the class (30–40) Therefore, the median class is (30–40) Using the formula: \[ \text{Median} = l + \left[\frac{\frac{n}{2} - cf}{f}\right] \times h \] Where: \(l = 30\), \(n = 70\), \(cf = 29\), \(f = 13\), \(h = 10\) \[ = 30 + \left[\frac{35 - 29}{13}\right] \times 10 = 30 + \frac{6}{13} \times 10 = 30 + \frac{60}{13} = 30 + 4.62 = 34.62 \text{ (approx.)} \]