â€ś Statistics is a numerical statement of facts in any department of enquiry placed in relation to each otherâ€™. -Bowley
â€ś Statistics are the classified facts representing the conditions of the people in a State specially those facts which can be stated in numbers or any tabular or classified arrangementâ€ť. -Webster
â€ś Statistics can be defined as the aggregate of facts affected to a marked extent by multiplicity of causes, numerically expressed, enumerated or estimated according to a reasonable standard of accuracy, collected in in a systematic manner, for a pre-determined purpose and placed in relation to each otherâ€ť. -Secrist
Statistics is the science of collecting, organizing , analyzing, interpreting and presenting data.
-Extensive use of Differentiation, Algebra, Trigonometry, Matrices etc in modern business analysis.
-Statistics now treated as Applied Mathematics.
4. Economics
- Family Budgeting
-Applied in solving economic problems related to production, consumption, distribution of products as per income & wealth related patterns, wages, prices, profits & individual savings, investments, unemployment & poverty etc.
Marketing Policy Decisions depend on forecasting, demand analysis, time & motion studies, inventory control, investments & analysis of consumer data for production & sales.
Cumulative Frequency of a class is the sum of the frequency of that class and the frequencies of all the preceding or succeeding classes which are listed in some sensible order (numerical order, alphabetical order, etc.)
f 4 f 4 x 4 x 4 f 3 f 3 x 3 x 3 f 2 f 2 x 2 x 2 f 1 f 1 x 1 x 1 Freq. fx X
34.
ÎĽ = 3144 / 50 = 62.88 Ans. Illustration 4 272 50 = N 3144 = ÎŁ fx 68 6 396 66 10 640 64 18 1116 62 12 60 x 12 = 720 60 No. of Students f fX Height (in inches) X
Q1.The avg. marks secured by 50 students was 44.Later on it was discovered that a score 36 was misread as 56. Find the correct average marks secured by the students.
Suppose for k different series with n 1 ,n 2 â€¦â€¦n k observations each, the respective A.M s are ÎĽ 1 , ÎĽ 2 ,â€¦. ÎĽ k . Then the A.M of the new series obtained on combining all the n 1 ,n 2 ,â€¦n k observations is obtained using the formula:
374 Less than 1000 392 Less than 1100 400 Less than 1200 324 Less than 900 265 Less than 800 194 Less than 700 116 Less than 600 60 Less than 500 20 Less than 400 0 Less than 300 No. of tubes Lifetime (in hrs.)
Being based on all the observations, is considerably affected by abnormal observations. For ex. A.M of 1000, 25, 35 & 40 will be (1000+25+35+40)/4 = 275 which is not at all a representative figure.
Cannot be calculated even if a single observation is missing.
Cannot be obtained just by inspection as in case of median & mode.
May give absurd results. For ex. If avg. no. of children per family is to be calculated and the result is 3.4 children per family, how would you interpret it?
(iii) Find the cum.freq. just greater than N/2. Suppose it is C.
(iv) Find the corresponding value of X. (the item) This is median.
62.
Calculation of Median-Illustration (Discrete Freq. Distribution) Here N = 50 (i) N/2 = 25 (ii) Cum. Frequency just greater than N/2 = 30 (iii)Corresponding value of item is 62. Median = 60 Ans. 12 12 60 30 18 62 40 10 64 46 6 66 50 4 68 N = 50 Cum. Freq. No. of students Height (in inches)
(iii) Find the cum.freq. just greater than N/2. Suppose it is X.
(iv) Look for the cum.freq. preceding X. Find the corresponding class interval.This is median class
Formula Used
Where L1 = L.L of median class
L2 = U.L of median class
C =cum.freq. of class preceding the median class.
f = frequency of median class.
Md = L1 + N/2 - C (L2 â€“ L1) f
64.
Calculation of Median-Illustration (Grouped Freq. Distribution) N/2 = 3600/2 = 1800 Cum.freq. just greater than 1800 is 2600. Hence median class is 25-30. Hence L1 = 25 L2 = 30 C = 1800 f = 800 Md = 25 + 1800 - 1800 (30 â€“ 25 ) 800 = 25 Ans. ÎŁ f= 3600 3600 400 35-40 3200 600 30-35 2600 800 25-30 1800 900 20-25 900 700 15-20 200 200 10-15 Cum. Freq Freq.(f) C.I
65.
Calculation of Missing Frequencies when median is known : Illustration : Median = 50 N = 100 15 56 + f1 + f2 80-100 ? = f 2 41+ f 1 +f 2 60-80 27 41 + f 1 40-60 ? = f 1 14 + f 1 20-40 14 14 0-20 No. of Families Cumulative Freq. Expenditure
66.
Calculation of Missing Frequencies when median is known : Illustration
Here median = 50 L 1 = 40
N = 100 L 2 = 60
N/2 = 50 f = 27
Hence median class 40-60 C = 14 + f1
Md = L 1 + N/2 - C (L 2 â€“ L 1 ) f 50 = 40 + 50 â€“ (14 + f 1 ) (60 â€“ 40) 27 10 = 720 â€“ 20 f 1 27 f 1 = 450/20 = 22.5 = 23 families approx. N = 56 + f 1 + f 2 100 = 56 + 23 + f 2 f 2 = 21 Ans. f 1 = 23 and f 2 = 21
7 55-60 13 50-55 15 45-50 20 40-45 30 35-40 33 30-35 28 25-30 14 20-25 No. of Persons Age 125 Less than 80 120 Less than 70 112 Less than 60 96 Less than 50 76 Less than 40 40 Less than 30 16 Less than 20 4 Less than 10 Frequency Value
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