CBSE Class 10 Mathematics Statistics Notes Set 07

Mean of Ungrouped and Grouped Data

Mean of n observations: \( \bar{x} = \frac{\sum x_i}{n} \). With frequencies (direct method): \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \).
Assumed mean method: take a convenient value A and deviations \( d_i = x_i - A \); then \( \bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} \). Step deviation method: with class width h and \( u_i = \frac{x_i - A}{h} \), \( \bar{x} = A + h\,\frac{\sum f_i u_i}{\sum f_i} \).
For a grouped table, use the class mark (mid-value) as x. Class mark = (upper limit + lower limit)/2. For inclusive classes such as 30-34, 35-39, the true class boundaries are 29.5-34.5, 34.5-39.5 and so on, and the class size is the difference of the boundaries. If the mean is given and one frequency is missing, write the mean formula and solve the equation for the unknown.

Median and Mode of Grouped Data

Median: form the cumulative frequency table, find N/2, and pick the class whose cumulative frequency first exceeds N/2 (the median class). Then \( \text{Median} = l + \left(\frac{N/2 - cf}{f}\right)h \), where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h the class width.
Mode: the modal class is the class with the highest frequency. \( \text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\,h \), where f1 is the modal class frequency and f0, f2 the frequencies of the classes before and after it.
Empirical relation for moderately skewed data: \( \text{Mode} = 3\,\text{Median} - 2\,\text{Mean} \).

Cumulative Frequency and Graphs

A less-than cumulative table adds frequencies from the lowest class upwards and is plotted against the upper class limits; a more-than table subtracts frequencies from the total and is plotted against the lower class limits. Both curves are ogives and meet at the median.
A histogram uses bars whose areas are proportional to the frequencies, so for equal class widths the heights are proportional to the frequencies. A bar diagram is used for discrete data and a frequency polygon joins the mid-points of the tops of the bars.
A change of the largest observations does not change the median, while the mean is affected by every observation, especially extreme ones. Adding an observation equal to the mean leaves the mean unchanged; adding 0 and the mean m to n observations of mean m gives a new mean (n+1)m/(n+2).

Worked Examples

Example 1: values 4, 6, 9, 10, 15 with frequencies 5, 10, 10, 7, 8. Σf = 40 and Σfx = 360, so the mean is 9.
Example 2: for the classes 0-10 to 40-50 with frequencies 7, 10, 15, 8, 10 take A = 25 and h = 10. Then Σfu = 4 and the mean is 25 + 10 × 4/50 = 25.8.
Example 3: marks (cumulative) below 10, 20, ..., 60 are 6, 15, 29, 41, 60, 70. N = 70, N/2 = 35, median class 30-40: median = 30 + 10 × (35 − 29)/12 = 35.