ch2measuresofcentraltendency-3.studentview

# ch2measuresofcentraltendency-3.studentview - Chapter 2...

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Unformatted text preview: Chapter 2: Chapter Measures of Central Tendency Measures To study variability, we first ignore variability, by looking at the “centre” of the data. There are several ways of describing the “centre”. A. The Mean Example 1: The salaries of five employees in company A are: 15.2, 9.2, 6.7, 12.4, 20.5 ( in \$1000s per month) Company A data: Company A data: 15.2 , 9.2 , 6.7 , 12.4 , 20.5 (\$1000s) Q: What is the mean salary of these five persons? Solution: Mean is µ = µA = 15.2 + 9.2 + 6.7 + 12.4 + 20.5 64 = = 12.8 = \$12,800 5 5 Notation Each problem involves one variable, here, salary. We denote it by x. 1st datum: x1 = 15.2 2nd datum: x2 = 9.2 etc….. Generally, a datum is represented by xi Now, x1 ,..... Now, xi subscript, index running subscript Total number of data = N (=5, here). Formula: x + x + ....... + x N µ= 1 2 = i =1 N N ∑x N i 1 = N ∑x i =1 N i [ = mu, Greek for “m”] µ [ = sigma, the Greek S, for summation] ∑ xi = summation of for i =1 to i = N i i =1 ∑ N x Sigma notation exercises: Sigma notation exercises: • Q1. Write each of the following in full, i.e. without the summation sign ∑ (3 y i=2 4 i − 2) • Q2. ∑ 3y i =2 4 i −2 • Solution: • Solution: ∑ (3 y i =2 4 i − 2) ∑3y i =2 4 i −2 = (3 y2 − 2) + (3 y3 − 2) + (3 y4 − 2) = 3 y 2 + 3 y3 + 3 y 4 − 6 = 3 y 2 + 3 y3 + 3 y 4 − 2 Sigma notation exercises cont’d: Sigma notation exercises cont’d: • Q3. Write each of the following in full, i.e. without the summation sign ∑ (3 y i =2 4 i − 2i ) • Solution: ∑ (3 y i =2 4 i − 2i ) = (3 y2 − 2 × 2) + (3 y3 − 2 × 3) + (3 y4 − 2 × 4) = (3 y2 − 4) + (3 y3 − 6) + (3 y4 − 8) = 3 y2 + 3 y3 + 3 y4 − 18 Example 2 : The following are the salaries of 6 employees of Company B: 8.7, 10.7, 7.3, 16.5, 21.8, 55.0. Q: Find the mean. Solution: µ = µB ∑x = 8.7 + 10.7 + ............ + 55 = = 20 = \$20,000 N 6 i Calculator steps for data entry Calculator steps for data entry ON MODE COMP (1) MODE SD (4) 8.7 DT 10.7 DT 7.3 DT 16.5 DT 21.8 DT 55 DT n SHIFT S­SUM (3) EXE SHIFT S­VAR (1) EXE x 8.7, 10.7, 7.3, 16.5, 21.8, 55.0 6 20 N =6 µ = 20 B: The Median B: The Median When data are arranged in their orders of magnitude (from smallest to the largest), the value in the middle is called the median of the data. N + 1 th ) score 2 Example 3: Find the median salary for each of the companies, A and B of the previous examples. Formula: Med = ( Solution: For company A, the data are arranged as: 6.7, 9.2, 12.4, 15.2, 20.5 Here, N=5, so the when ranked in order 5 +1 Med = ( ) = 3rd score Therefore, Median = 12.4=\$12,400 2 For B, the data are arranged as: 7.3, 8.7, 10.7, 16.5, 21.8, 55.0 Now there are two numbers, 10.7 and 16.5, in the middle, (the 3rd and 4th scores). We take their average as the median. ∴ Med = 1 rd 10.7 + 16.5 (3 + 4th ) scores = = 13.6 = \$13,600 2 2 Med = ( N + 1 th 6 +1 ) score = ( ) = 3.5th score 2 2 Note 1: There are slightly different ways for treating odd N and even N, for ungrouped data. [For grouped data, however, we need NOT bother about the oddness or evenness of N. See Chapter 5 E.] Note 2: Note 2 For cases which contain some “extreme” values (such as 55 in company B), the median is often a more useful measure of central tendency. 13.6 20 ___|__|____|__|__|____|_|__________________|___ 7.3 8.7 10.7 16.5 21.8 55 The median is said to be a “ROBUST statistic”, which means that it is NOT effected by “extreme values” or “outliers”. The mean , however, is more popularly used in general, as it makes use of all information available. It’s easier to handle. Note 3: Notably, in problems concerning age or salary, the use of median is recommended. C: The Mode C: The Mode Example 4: The age distribution of a class of eighteen first year students are given: 18, 19, 17, 20, 19, 18, 19, 19, 20, 19, 18, 18, 21, 17, 19, 19, 18, 17. Q: Find the mode (i.e. the modal age) Solution: The value that appears most frequently is the mode. age freq 17 3 18 5 19 7 20 2 21 1 total 18 Here the highest frequency is 7. Therefore Mode = 19 yrs Calculator steps for MULTIPLE data Calculator steps for MULTIPLE data age freq 17 3 18 5 19 7 20 2 21 1 total 18 ON MODE COMP (1) MODE SD (4) 17 SHIFT 3 DT 18 SHIFT ; 5 DT 19 SHIFT ; 7 DT 20 SHIFT ; 2 DT 21 DT n 18 SHIFT S­SUM (3) EXE SHIFT S­VAR (1) EXE 18.6111 x , ; N =18 µ = 18.611111 D: Geometric Mean D: Geometric Mean Example 5: A sum of money, invested in a certain fund, grew, in 4 years, by 25%, 12%, ­ 4% and 10%. What is the average growth rate per annum? Solution: Solution Let A=the initial amount of money to be invested By the end of 4 years, it becomes B=A(1+0.25)(1+0.12)(1 ­ 0.04)(1+0.10) = 1.4784 A Let r*= the average annual growth rate. Then we want r* to satisfy the relation: B = A(1 + r * )(1 + r * )(1 + r * )(1 + r * ) = A(1 + r * ) 4 ∴(1 + r * ) 4 =1.4784 1 + r * = 4 1.4784 ∴r * = 0.1027 =10.27% RULE: B = A(1 + r*) RULE: An = P (1 + r ) n You may remember (from high school) the formula for calculating Compound Interest: n These are the same (except for the notation) “The total amount of money that you have at the end of n years is equal to the total Principal Amount invested multiplied by (1+rate) raised to the power of n” (where n=the no of years) With calculator press: With 1.25 x 1.12 x 0.96 x 1.10 4 SHIFT x EXE 1.4784 1.102676 ∧ -1 HEX Ans EXE EXE 0.102676 The geometric mean is 0.1027 = 10.27% The Note: Let the individual rates be Note 1 r1 , r2 , r3 , r4 Then the geometric mean of (1 + r ), (1 + r ), (1 + r ), (1 + r ) is denoted by 1+r*, which is given by 2 3 4 (1 + r * ) 4 = (1 + r1 )(1 + r2 )(1 + r3 )(1 + r4 ) In general, the geometric mean of y1 , y2 , y3 ,........., y N is N y1 y2 y3 ......... y N Example 6: Example Q: Find the geometric mean of Find 3.2, 5.7, 0.9, 4.3 and 3.4 Solution: The geometric mean is 5 3.2 × 5.7 × 0.9 × 4.3 × 3.4 = 5 240.00192 = 2.992560528 With calculator press: 3.2 x 5.7 x 0.9 x 4.3 x 3.4 .2 5 SHIFT x EXE 240.00192 2.992560528 Ans EXE The geometric mean is 2.99256 The Example 7: Example 7: Q: Fund A was worth \$2.5m in 2000. It grew by 8%, ­3%, ­5%, 2% and 9% in the subsequent 5 years. Fund B was worth \$3.5 m in 2000. It grew by 2%, 4%, ­10%, 3% and 18% in the subsequent 5 years. What was the average annual growth rate for the two funds taken together? Solution: Solution Let A and B be the principal amounts for fund A and fund B. A = \$2.5 m , B = \$3.5 m , n = 5 XB XA Let and , be the final values for fund A and B respectively. XA = 2.5(1.08)(0.97)(0.95)(1.02)(1.09) = 2.76621399 XB = 3.5(1.02)(1.04)(0.90)(1.03)(1.18) = 4.061283408 Let r be the average growth rate then : ( A + B )(1 + r ) n = ( X A + X B ) ( 2.5 + 3.5)(1 + r ) 5 = ( 2.76621399 + 4.061283408) 6.827497398 (1 + r ) = 6 5 therefore r = 2.62% 1 + r = 5 1.137916233 With calculator press: With 2.5 x 1.08 x 0.97 x 0.95 x 1.02 x 1.09 EXE 2.5 2.76621399 6.827497398 + 3.5 x 1.02 x 1.04 x 0.90 x 1.03 x 1.18 3.5 EXE ÷ 6 EXE x 1.137916233 5 SHIFT -1 Ans EXE 1.026176485 EXE 0.026176485 The geometric mean is 0.0261765 = 2.62% The ...
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