| --- |
| source: ncert |
| topic: statistics |
| class_level: class_9 |
| chapter: ch14 |
| difficulty: intermediate |
| --- |
| |
| Class 9 | Ch14: Statistics |
| TYPES OF DATA: |
| Primary data: collected by investigator. Secondary data: from other sources. |
| Raw data: unorganised. Grouped data: organised in class intervals. |
| FREQUENCY DISTRIBUTION TABLE: |
| Class interval: group of values (0-10, 10-20 etc). Class width=upper-lower limit. |
| Frequency: number of observations in each class. |
| Class mark (midpoint)=(upper limit+lower limit)/2. |
| Cumulative frequency: running total of frequencies. |
| MEAN FOR GROUPED DATA: |
| Direct method: Mean = sum(f×x)/sum(f) where x=class mark, f=frequency. |
| Assumed mean method: Mean = a + sum(f×d)/sum(f) where d=x-a, a=assumed mean. |
| Step deviation method: Mean = a + (sum(f×u)/sum(f))×h where u=(x-a)/h, h=class width. |
| MEDIAN FOR GROUPED DATA: |
| Median = L + ((n/2-cf)/f)×h |
| where L=lower limit of median class, n=total frequency, cf=cumulative frequency before median class, f=frequency of median class, h=class width. |
| MODE FOR GROUPED DATA: |
| Mode = L + ((f1-f0)/(2f1-f0-f2))×h |
| where L=lower limit of modal class, f1=frequency of modal class, f0=previous class frequency, f2=next class frequency, h=class width. |
| EMPIRICAL RELATIONSHIP: Mode = 3×Median - 2×Mean (approximately). |
| SOLVED EXAMPLES: |
| Example 1: Mean by assumed mean. Class marks: 10,20,30,40,50. Frequencies: 4,6,8,5,3. Assumed mean=30. |
| d values: -20,-10,0,10,20. sum(f×d)=4(-20)+6(-10)+8(0)+5(10)+3(20)=-80-60+0+50+60=-30. |
| Mean=30+(-30/26)=30-1.15=28.85. |
| Example 2: Find median class for n=26. n/2=13. Cumulative frequencies: 4,10,18,... |
| 13 falls in third class (cf=10, f=8). Median class is the third class interval. |
| COMMON MISTAKES: |
| Class mark is MIDPOINT of class interval, not lower or upper limit. |
| For median: find n/2 first, then locate which cumulative frequency it falls in. |
| Modal class has HIGHEST frequency, not highest value. |
