Unit 5: Statistics
એકમ ૫: આંકડાશાસ્ત્ર
5.1 Mean for ungrouped and grouped data
Definition
The Arithmetic Mean is the average of a given set of data. For ungrouped data, it is the sum of all observations divided by the number of observations. For grouped data, frequencies are used as weights.
Real-Life Application: Calculating the average test scores of polytechnic students to evaluate overall class performance.વ્યાખ્યા
સમાંતર મધ્યક એ આપેલી માહિતીની સરેરાશ છે. અવર્ગીકૃત માહિતી માટે, તે બધા અવલોકનોના સરવાળાને કુલ સંખ્યા વડે ભાગીને મેળવવામાં આવે છે. વર્ગીકૃત માહિતીમાં આવૃત્તિનો ઉપયોગ થાય છે.
વ્યવહારુ ઉપયોગ: વર્ગના એકંદર પ્રદર્શનનું મૂલ્યાંકન કરવા માટે પોલિટેકનિકના વિદ્યાર્થીઓના સરેરાશ ગુણની ગણતરી કરવી.Example 1: Ungrouped Data (અવર્ગીકૃત માહિતી)
Question: Find the mean of: 5, 8, 12, 15, 20. (મધ્યક શોધો.)
Solution (ઉકેલ): Here, \( n = 5 \)| Observation \( (x_i) \) | 5 | 8 | 12 | 15 | 20 | Total \( \sum x_i = 60 \) |
|---|
\( \overline{x} = \frac{\sum x_i}{n} = \frac{60}{5} = 12 \)
Answer: \( \overline{x} = 12 \)
Example 2: Discrete Grouped Data (અસતત આવૃત્તિ વિતરણ)
Question: Find the mean. (મધ્યક શોધો.)
Solution (ઉકેલ):| \( x_i \) | \( f_i \) | \( f_i x_i \) |
|---|---|---|
| 10 | 2 | 20 |
| 20 | 5 | 100 |
| 30 | 2 | 60 |
| 40 | 1 | 40 |
| Total | \( N = 10 \) | \( \sum f_i x_i = 220 \) |
\( \overline{x} = \frac{\sum f_i x_i}{N} = \frac{220}{10} = 22 \)
Answer: \( \overline{x} = 22 \)
Example 3: Continuous Grouped Data (વર્ગીકૃત માહિતી - Step Deviation)
Question: Find the mean. Class: 0-10, 10-20, 20-30, 30-40. Freq: 4, 6, 8, 2.
Solution (ઉકેલ): Let Assumed Mean \( A = 15 \) (Mid-value of 10-20), Class length \( c = 10 \).| Class | Mid-value \( (x_i) \) | Freq \( (f_i) \) | \( d_i = \frac{x_i - 15}{10} \) | \( f_i d_i \) |
|---|---|---|---|---|
| 0-10 | 5 | 4 | -1 | -4 |
| 10-20 | 15 (A) | 6 | 0 | 0 |
| 20-30 | 25 | 8 | 1 | 8 |
| 30-40 | 35 | 2 | 2 | 4 |
| Total | \( N = 20 \) | \( \sum f_i d_i = 8 \) | ||
\( \overline{x} = A + \left(\frac{\sum f_i d_i}{N}\right) \times c \)
\( \overline{x} = 15 + \left(\frac{8}{20}\right) \times 10 = 15 + 4 = 19 \)
Answer: \( \overline{x} = 19 \)
Example 4: Missing Frequency (ખૂટતી આવૃત્તિ)
Question: Mean is 22. Find missing frequency k. (મધ્યક 22 છે. k શોધો.)
Solution (ઉકેલ):| \( x_i \) | \( f_i \) | \( f_i x_i \) |
|---|---|---|
| 10 | 2 | 20 |
| 20 | k | 20k |
| 30 | 3 | 90 |
| 40 | 1 | 40 |
| Total | \( \sum f_i = 6 + k \) | \( \sum f_i x_i = 150 + 20k \) |
\( \overline{x} = \frac{\sum f_i x_i}{N} \Rightarrow 22 = \frac{150 + 20k}{6 + k} \)
\( 132 + 22k = 150 + 20k \Rightarrow 2k = 18 \Rightarrow k = 9 \)
Answer: \( k = 9 \)
5.2 Mean deviation and Standard deviation about Mean
Definition
Mean Deviation: The mean of the absolute deviations of observations from the arithmetic mean.
Standard Deviation: The square root of the variance, measuring the amount of variation or dispersion in a set of values.
વ્યાખ્યા
સરેરાશ વિચલન: મધ્યકથી લીધેલા અવલોકનોના તફાવતોના નિરપેક્ષ મૂલ્યોની સરેરાશ.
પ્રમાણિત વિચલન: એ માહિતીના પ્રસારનું માપ દર્શાવે છે (વિચરણનું વર્ગમૂળ).
Example 1: Standard Deviation - Ungrouped (અવર્ગીકૃત માહિતી)
Question: Find S.D. of 2, 4, 6, 8, 10. (પ્રમાણિત વિચલન શોધો.)
Solution (ઉકેલ): First, find Mean \( \overline{x} = \frac{30}{5} = 6 \).| \( x_i \) | \( x_i - \overline{x} \) | \( (x_i - \overline{x})^2 \) |
|---|---|---|
| 2 | -4 | 16 |
| 4 | -2 | 4 |
| 6 | 0 | 0 |
| 8 | 2 | 4 |
| 10 | 4 | 16 |
| Total | 0 | \( \sum = 40 \) |
\( \sigma = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n}} = \sqrt{\frac{40}{5}} = \sqrt{8} \approx 2.828 \)
Answer: \( \sigma \approx 2.828 \)
Example 2: Standard Deviation - Grouped (વર્ગીકૃત માહિતી)
Question: Find S.D. Class: 0-10, 10-20, 20-30. Freq: 1, 2, 1.
Solution (ઉકેલ): Let Assumed Mean \( A = 15 \), Class length \( c = 10 \).| Class | \( x_i \) | \( f_i \) | \( d_i = \frac{x_i-15}{10} \) | \( f_i d_i \) | \( f_i d_i^2 \) |
|---|---|---|---|---|---|
| 0-10 | 5 | 1 | -1 | -1 | 1 |
| 10-20 | 15 | 2 | 0 | 0 | 0 |
| 20-30 | 25 | 1 | 1 | 1 | 1 |
| Total | \( N = 4 \) | \( \sum f_i d_i = 0 \) | \( \sum f_i d_i^2 = 2 \) | ||
\( \sigma = \sqrt{\frac{\sum f_i d_i^2}{N} - \left(\frac{\sum f_i d_i}{N}\right)^2} \times c \)
\( \sigma = \sqrt{\frac{2}{4} - 0^2} \times 10 = \sqrt{0.5} \times 10 = 0.707 \times 10 = 7.07 \)
Answer: \( \sigma = 7.07 \)
Example 3: Mean Deviation - Discrete Data (સરેરાશ વિચલન)
Question: Find Mean Deviation about Mean. \( x_i \): 2, 4, 6. \( f_i \): 3, 4, 3.
Solution (ઉકેલ):| \( x_i \) | \( f_i \) | \( f_i x_i \) | \( |x_i - \overline{x}| \) | \( f_i |x_i - \overline{x}| \) |
|---|---|---|---|---|
| 2 | 3 | 6 | 2 | 6 |
| 4 | 4 | 16 | 0 | 0 |
| 6 | 3 | 18 | 2 | 6 |
| Total | \( N = 10 \) | \( \sum = 40 \) | \( \sum = 12 \) |
Mean \( \overline{x} = \frac{40}{10} = 4 \).
\( M.D. = \frac{\sum f_i |x_i - \overline{x}|}{N} = \frac{12}{10} = 1.2 \)
Answer: M.D. = 1.2
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