STATISTICS STUDY OTES UIT I MEASURES OF CETRAL TEDECY IDIVIDUAL SERIES ARITHMETIC MEA: Direct Method X X Short-cut Method X A d Step-Deviation Method X A d i MEDIA: th Size of term MODE: Either by inspection or the value that occurs largest number of times EMPIRICAL RELATIO: DISCRETE SERIES Direct Method X f X Short-cut Method X A f d Step-Deviation Method X A f d i Size of th term Grouping Method determines that value around which most of the frequencies are concentrated. Mode = 3 Mean Median COTIUOUS SERIES Direct Method X f X Short-cut Method X A f d Step-Deviation Method X A f d i th Size of term Median = L c. f. i Mode = L f f 0 i f (f f ) 0 Problems. From the following data compute Arithmetic Mean Marks 0-0 0-0 0-30 30-40 40-50 50 60 o. of students 5 0 5 30 0 0 Marks Midvalue o. of students f x X f 0-0 5 5 5
0-0 5 0 50 0-30 5 5 65 30-40 35 30 050 40-50 45 0 900 50-60 55 0 550 =00 3300
Arithmetic Mean X f X 3300 33 00. Calculate Arithmetic Mean from the following data Marks 0-0 0-30 30-60 60-00 o. of students 5 5 8 The class intervals are unequal but still to simplify calculations we can take 5 as common factor. Marks Midvalue o. of students d f d x f (x - 45) / 5 0 0 5 5-8 - 40 0 30 5-5 - 60 30 60 5 5 0 0 60-00 35 8 7 56 = 50-44 Arithmetic Mean X A f d i A = 45, f d= - 44, = 50, I = 5 X 45 44 5 45 4.4 40.6 50 3. Find the missing frequency from the following data Marks 0 0 0 0 0-30 30-40 40-50 50 60 o. of Students 5 5 0-0 0 The arithmetic mean is 34 marks. Let the missing frequency be denoted by X
Marks Midvalue f` f x x 0 0 5 5 5 0 0 5 5 5 0 30 5 0 500 30 40 35 X 35X 40-50 45 0 900 50 60 55 0 550 = 70 + X 00+35X X f x 34 00 35 X 70 X 34 (70 + X) = 00 + 35X 380 + 34X = 00 + 35X 35X 34X = 380 00 X = 80 4. Calculate the Median and Mode from the following data Central size 5 5 35 45 55 65 75 85 Frequencies 5 9 3 0 5 8 3 Since we are given central values first we determine the lower and upper limits of the classes. The class interval is 0 and hence the first class would be 0 0. Class Midvalue f d f d c.f. x (x 55) / 0 0-0 5 5-4 - 0 5 0-30 5 9-3 - 7 4 30-40 35 3 - - 6 7 40-50 45 - - 48 50-60 55 0 0 0 68 60-70 65 5 5 83 70-80 75 8 6 9 80-90 85 3 3 9 94 fd = - 54 Calculation of Median: Med = size of th term = 94 47
Median lies in the class 40 50 Median = L / c. f. i f M 40 47 7 0 40 9.54 = 49.54 5. Calculate the median and mode of the data given below. Using then find arithmetic mean Marks 0-0 0 0 0-30 30-40 40-50 50 60 o. of Students 8 3 45 65 75 80 Marks f` c.f. 0 0 8 8 0 0 5 3 0 30 45 30 40 0 65 40 50 0 75 50 60 5 80 = 80 Calculation of Median: Med = size of th term = 80 40 th item Median lies in the class 0 30 Median = L / c. f. i f M 0 40 3 0 0 7.73 = 7.73 Mode lies in the class is 0 30 f f 0 5 Mode = L i 0 0 7.78 f (f f ) 44 (5 0) 0
MEASURES OF DISPERSIO IDIVIDUAL OBERSERVATIOS DISCRETE& COTIUOUS SERIES QUARTILE DEVIATIO: Q.D. = Q 3 Q Coefficient of Q.D. = Q Q 3 Q Q 3 Quartile Deviation: Q.D. = Q Q 3 Coefficient of Q.D. = Q 3 Q Q Q 3 STADARD DEVIATIO: Actual Mean Method: Actual Mean Method: f ( X X ) f ( X X ) Assumed Mean Method: d d Step Deviation Method Assumed Mean Method: fd fd Step Deviation Method d d i fd fd i C.V. 00 X C.V. 00 X. Find the Mean and standard deviation from the following distribution Mid value.0.5 3.0 3.5 4 4.5 5 5.5 6 o. of Students 6 36 60 76 37 8 3
Midvalue o. of Students d f d f d x f (x 4) / 0.5.0-4 - 8 3.5 6-3 - 48 44 3 36 - - 7 44 3.5 60 - - 60 60 4 76 0 0 0 4.5 37 37 37 5 8 36 7 5.5 3 3 9 7 6.0 4 8 3 f = 50 fd = - 98 d 548 Mean X A f d i 4 98 0.5 3.8 50 Standard deviation fd fd i 548 98.05 0.75 50 50. Find the Standard deviation and Coefficient of Variation from the following data Marks o. of students Up to 0 Up to 0 30 Up to 30 65 Up to 40 07 Up to 50 57 Up to 60 0 Up to 70 Up to 80 30 Class Midvalue o. of Students d f d f d x f (x 35) / 0 0 0 5-3 - 36 08 0-0 5 8 - - 36 7 0-30 5 35 - - 35 35 30-40 35 4 0 0 0 40-50 45 50 50 50
50-60 55 45 90 80 60-70 65 0 3 60 80 70-80 75 8 4 3 8 = 30 fd =5 fd 753
Mean X A f d i 35 5 0 40.43 30 Standard deviation fd fd i 753 5 0 7.6 30 30 C.V. 00 7.6 00 4.69 X 40.43 3. The scores of two batsmen A and B in ten innings during a certain season are: A 3 8 47 63 7 39 0 60 96 4 B 9 3 48 53 67 90 0 6 40 80 Find which of the two batsmen more consistent in scoring X X X X X Y Y Y Y Y 3-4 96 9-3 96 8-8 34 3-9 36 47 48-4 63 7 89 53 3 9 7 5 65 67 7 89 39-7 49 90 40 600 0-36 96 0-40 600 60 4 96 6 44 96 50 500 40-0 00 4-3 04 80 30 900 X = 460 Batsman A: Y = 500 X X = 6500 Y Y = 5968 Mean X X 460 0 46 (X 6500 X) 5.495 0
C.V. 00 5.495 00 55.4 X 46 Batsman B: Mean Y Y 500 50 0 Y Y 5968 4.43 0 00 C.V. 00 4.43 48.86 Y 50 Since Coefficient of Variation is less in the case of Batsman B, we conclude that the Batsman B is more consistent. 4. Calculate the Quartile deviation and the coefficient of quartile deviation from the following data Marks o. of students Below 0 8 Below 40 0 Below 60 50 Below 80 70 Below 00 80 Marks f` c.f. 0-0 8 8 0-40 0 40-60 30 50 60-80 0 70 80-00 0 80 = 80 Q is the size of / 4 th item. Q lies in the class 0 40 Q L / 4 c.f. i 0 0 8 0 40 f Q3 is the size of 3 / 4 th item. Q3 lies in the class 60 80
Q L 3 / 4 c. f. i 60 60 50 0 70 3 f 0 Q.D. Q Q 70 40 3 5 Q Q 30 3 Coefficient of QD Q Q 0 0.73 3 5. Calculate the Inter-Quartile range and the coefficient of quartile deviation from the following data Marks o. of students Above 0 50 Above 0 40 Above 0 00 Above 30 80 Above 40 80 Above 50 70 Above 60 30 Above 70 4 Above 80 0 Marks f` c.f. 0-0 0 0 0-0 40 50 0-30 0 70 30-40 0 70 40-50 0 80 50-60 40 0 60-70 6 36 70-80 4 50 = 50 Q is the size of / 4 th item. Q lies in the class 0 0 Q L / 4 c.f. i 0 37.5 0 0 6.875 f 40 Q3 is the size of 3 / 4 th item. Q3 lies in the class 50 60 Q 3 L 3 / 4 c.f. i 50.5 80 0 58.5 f 40
Inter Quartile Range Q Q 4.375 3 Coefficient of QD Q Q 4.375 3 0.55 Q Q 75 3 MOMETS Moments about mean (X X) 0 (X X) 3 4 (X X) (X X) 3 4 In a Frequency distributiom f (X X) 0 f (X X) Moments about arbitrary origin (X A) ( X A) In a frequency distribution f (X A) f ( X A ) Moments about mean 3 f (X X) 3 4 f (X X) 4 3 (X A)3 4 (X A) f (X A) 3 3 4 f (X A) 4 4 4 43 6 3 44, 3 3 3 3
SKEWESS AD KURTOSIS Karl Pearson s Skewness = Mean - Mode Bowley s Skewness = Q3 + Q- Med Mean Mode Karl Pearson s coefficient of Skewness = Bowley s coefficient of Skewness = Q 3+ Q - Med Q 3 - Q, 3 4 3 3, 3 3 3. Calculate the coefficient of skewness by Karl Pearson s method and the values of β and β from the following data Profits o. of (in lakhs) companies 0-0 8 0-30 0 30-40 30 40 50 50 60 0 Solution : Class Midvalue o. of d f d f d f d 3 f d 4 x Students (x 35) f / 0 0-0 5 8 - - 36 7-44 88 0-30 5 0 - - 0 35-0 0 30-40 35 30 0 0 0 0 0 40-50 45 50 50-60 55 0 0 80 80 60 = 00 fd 4 fd 54 fd 3 6 fd 4 490
X A f d i 35 00 4 0 33.6
Modal class 30-40 Mode L f f 0 i 30 30 0 f (f f ) 60 (0 ) 0 35.56 fd fd i 54 4 0.33 00 00 Karl Pearson s coefficien t of Skewness = fd 0.4 f d.54 f d 3 0.6 3 f d 4 4.9 4 3 4.504, 3 3 4 43 3 0.03 3 Mean Mode 33.6 35.56 0.59.33 6 4 4.735
0.00045 3 0.0008 3 3.5458 4 4.735.3.048
. By measuring the quartiles find a measure of skewness for the following distribution Annual Sales o. of firms Less than 0 30 Less than 30 5 Less than 40 465 Less than 50 580 Less than 60 634 Less than 70 644 Less than 80 650 Less than 90 665 Less than 00 680 Sales f` c.f. 0-0 30 30 0-30 95 5 30-40 40 465 40-50 5 580 50-60 54 634 60-70 0 644 70-80 6 650 80-90 5 665 90-00 5 680 = 680 Q lies in the class 0-30 Q L / 4 c.f. i 0 70 30 0 7.8 f 95. Q3 lies in the class 40 50 Q L 3 / 4 c.f. i 40 50 465 0 43.9 3 f 5 Inter Quartile Range Q 3 Q 4.375 Coefficient of QD Q 3 Q 4.375 0.55 Q 3 Q 75 Median class 30-40 This is a SAMPLE (Few pages have been extracted from the complete notes:-it s meant to show you
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