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.