| --- |
| source: ncert |
| topic: statistics |
| class_level: class_11 |
| chapter: ch15 |
| difficulty: intermediate |
| --- |
| |
| Class 11 | Ch15: Statistics |
| MEASURES OF DISPERSION (how spread out data is): |
| Range = Maximum - Minimum. |
| MEAN DEVIATION: |
| Mean deviation about mean: MD(x_bar)=sum(|xi-x_bar|)/n. |
| Mean deviation about median: MD(M)=sum(|xi-M|)/n. |
| For grouped data: MD=sum(f|xi-x_bar|)/sum(f). |
| VARIANCE AND STANDARD DEVIATION: |
| Variance: sigma^2=sum((xi-x_bar)^2)/n (ungrouped). |
| For grouped: sigma^2=sum(f(xi-x_bar)^2)/sum(f). |
| Shortcut: sigma^2=(sum(xi^2)/n)-(x_bar)^2 = (sum(fi*xi^2)/sum(fi))-(x_bar)^2. |
| Standard Deviation sigma=sqrt(Variance). |
| COEFFICIENT OF VARIATION: CV=(sigma/x_bar)×100. |
| Used to compare variability of two distributions with different means. |
| Higher CV = more variable/inconsistent. |
| For comparing: data with higher CV is more variable. |
| ANALYSIS OF FREQUENCY DISTRIBUTIONS: |
| Two distributions compared by CV when means are different. |
| Two distributions compared directly by sigma when means are same. |
| SOLVED EXAMPLES: |
| Example 1: Find variance and SD of 6,8,10,12,14. |
| Mean=50/5=10. |
| Deviations from mean: -4,-2,0,2,4. Squared: 16,4,0,4,16. |
| Variance=40/5=8. SD=sqrt(8)=2sqrt(2)≈2.83. |
| Example 2: Two distributions: A has mean=25, SD=5. B has mean=10, SD=4. |
| CV(A)=(5/25)×100=20%. CV(B)=(4/10)×100=40%. |
| B is more variable (higher CV). |
| Example 3: Find MD about mean for 2,4,6,8,10. |
| Mean=6. |deviations|=4,2,0,2,4. MD=12/5=2.4. |
| COMMON MISTAKES: |
| Variance is always non-negative (it's sum of SQUARES divided by n). |
| SD=sqrt(Variance), NOT Variance=SD^2 (both are true, don't confuse direction). |
| CV is in PERCENTAGE: multiply by 100. |
