source: ncert
topic: statistics
class_level: class_10
chapter: ch14
difficulty: intermediate
Class 10 | Ch14: Statistics MEAN FOR GROUPED DATA: Direct method: x_bar = sum(fx)/sum(f). x=class mark=(upper+lower)/2. Assumed mean method: x_bar = a + sum(fd)/sum(f). d=x-a. Step deviation method: x_bar = a + (sum(f*u)/sum(f))×h. u=(x-a)/h. Best when h is constant. MEDIAN FOR GROUPED DATA: Formula: M = L + ((n/2 - cf)/f) × h. L=lower boundary of median class. n=total frequency. cf=cumulative frequency of class BEFORE median class. f=frequency of median class. h=class width. Median class: class where cumulative frequency first exceeds n/2. MODE FOR GROUPED DATA: Modal class=class with highest frequency. Mode = L + ((f1-f0)/(2f1-f0-f2)) × h. L=lower boundary of modal class. f1=frequency of modal class. f0=frequency of class BEFORE modal class. f2=frequency of class AFTER modal class. EMPIRICAL RELATIONSHIP: Mode = 3×Median - 2×Mean (approximately true). OGIVE (CUMULATIVE FREQUENCY CURVE): Less than ogive: plot (upper class boundary, cumulative frequency). More than ogive: plot (lower class boundary, more than cumulative frequency). Median from ogive: find n/2 on y-axis, draw horizontal to curve, then vertical to x-axis. SOLVED EXAMPLES: Example 1: Mean by step deviation. Classes 10-20,20-30,...50-60. Frequencies 4,6,8,5,3. a=30, h=10. Class marks: 15,25,35,45,55. u=(-2,-1,0,1,2). f×u=-8,-6,0,5,6. sum(f)=26. sum(f×u)=-3. Mean=30+(-3/26)×10=30-1.15=28.85. Example 2: Find median. n=26, n/2=13. cf values: 4,10,18,23,26. 13 falls in class 30-40 (cf=10, f=8). Median=30+((13-10)/8)×10=30+3.75=33.75. COMMON MISTAKES: For median formula: cf is cumulative frequency of class BEFORE (not including) median class. Class mark is MIDPOINT: for class 10-20, mark=15 (not 10 or 20). Modal class: highest frequency (not highest class mark).