ch6_some_techniques_of_calculation.studentview

# I2 y 9 3889 400 9 2 321111 z 964 2925 2 20 400 9 2

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Unformatted text preview: 6.7 , 2.4 6.7 N = 13 µ =8 -5.8 -5.8 3 9.7 σ 2 =14 Set 2 should be viewed as Set the this table: the x f 8 − 14 6.5 8 + 14 6.5 13 Solution: ON MODE COMP MODE SD 6.7 DT 2.4 DT 6.7 - 5.8 DT SHIFT 5.8 88+ SD 3 9.7 DT 14 ) EXE SHIFT ; 6.5 DT 14 ) EXE SHIFT ; 6.5 DT SHIFT SHIFT SHIFT n S-SUM EXE 17 = N S-VAR x EXE 6.4372 = µ S-VAR xσn EXE 4.841 = σ ∴ µ = 6.4372, σ = 4.8410 Simple Exercises: Simple 1. Create 10 numbers so that the mean=143 and 1. s.d.=15 s.d.=15 2. Data set 1: N1 = 10 µ1 = 141 σ1 =15 Data set 2: N 2 = 15 µ 2 = 156 σ 22 = 345 Data Find the mean and s.d. of the 25 numbers. B. Linear Transformation B. Consider the data set x1 , x2 ,......., xN Consider 2 µ X and σ X Suppose their mean and variance are respectively. Often we are interested in a linear transformation of them: yi = a + bxi ............(1) them: Thus, we have new data: y1 , y2 ,......., y N For instance x1 , x2 ,......., xN may be temperatures in For centigrade. Now we like to express them in Fahrenheit: Fahrenheit: y = 32 + 9 x i 5 i We like to ask: What are the mean µ and variance these new data y = a + bx ? these Y i The answers are: Proof: From (1), From yi = a + bxi N ∑y i =1 i = Na + b∑ xi i =1 N ∑y i =1 N i = a+b i µ =a +bµ .......... ..( 2) Y X 2 σ2 =b 2σX .......... ......(3) Y σ = b | × X .......... ....( 4) | σ Y µY = a + bµ X yi − µY = a + bxi − a − bµ X N N ∑x i =1 N N ∴ µ Y = a + bµ X i of 2 σY ∑(y − µ i =1 i N ∑(y − µ i =1 i Y Y) =b ∑(y − µ i =1 Y N 2 2 σ Y = b 2σ X 2 i =1 2 N i N ) = ∑ (a + bxi − a − bµ X ) = b 2 ) 2 N ∑ (x − µ i =1 i )2 X N 2 = b2 ∑ (x − µ i =1 i N X )2 2 N ∑ (x − µ i =1 i X )2 Example 6: Salaries (in \$1000 p.m.) of the Example Salaries employees of Company C are given below: employees Salary x Frequency f 5.5 6.5 8.0 total 20 50 25 5 (a) Find the mean and standard deviation of X. (b) If everyone’s salary is revised 6% up, find the (b) new mean and s.d. new (c) If everyone is further given a travel allowance (c) of \$400, find the new mean and s.d. of Solution: Solution (a) By use of SD-MODE, 5.5 SHIFT ; 20 DT 6.5 SHIFT ; 25 DT 8 SHIFT ; 5 DT N X = 50, µ X = 6.25, σ X = 0.75 we can easily obtain (b) By (1) (b) by (2) by by (4) (c) Now yi = 0 + 1.06 xi µY = 0 + 1.06 µ X = 1.06 × 6.25 = 6.625 2 2 σ Y = 1.06 2 σ X σ Y =| 1.06 | ×σ X = 1.06 × 0.75 = 0.795 wi = 0.4 + yi µW = 0.4 + µY = 0.4 + 6.625 = 7.025 2 2 σW = σY σ W = σ Y = 0.795 Example 7: The pass mark of an examination is 50. The table below shows the marks (X) of 250 students who took the examination. The examiner revised the marks by the formula given by Y. given Mark Mark X No. of Mean SD No. Mean SD students students Revised mark Y x ≥ 50 200 70 8 4 y = 60 + ( x − 50) 5 x < 50 50 40 6 11 y = 60 − (50 − x) 10 Find the mean and standard deviation of the 250 revised marks. revised Solution: Solution For the 200 students whose old mark are x ≥ 50 , 4 Mean(Y ) = 60 + (70 − 50) = 76 5 4 SD(Y ) = × 8 = 6.4 5 Similarly, for the 50 students with x < 50 , Similarly, Mean(Y ) = 60 − 11 (50 − 40) = 49 10 SD(Y ) = 11 × 6 = 6.6 10 Hence the mean & s.d. of the 250 new marks can be Hence found from the frequency table: found x 76-6.4= 69.6 82.4 42.4 55.6 55.6 f 100 100 25 25 76+6.4= 49-6.6= 49+6.6= ∴ µ = 70.6, σ = 12.5748 N 250 Simple Exercise: Simple Data set : x , x ,........, x Data Mean=12, Median=14, Mode=15 and CV=30%. Define new data set: yi = 80 − 4 xi i = 1,2,........,80 1 2 80 Find the mean, median, mode, s.d, CV and Sk for the yi ' s . Sk Solution: Solution RULES: RULES Y = a + cX then µY = a + cµ X MedY = a + c ⋅ Med X ModY = a + c ⋅ Mod X σ Y =| c | ⋅σ X and 2 2 σ Y = c 2 ⋅σ X Given and µX =12 C .V X = Med X =14 σX Mod X =15 σX =30% µX = 0.3 12 ∴ X =3.6 σ Y = a + cX Y = 80 − 4 X ⇒ a = 80, c = −4 µY = a + cµ X = 80 − 4 ×12 = 32 MedY = a + c ⋅ Med X = 80 − 4 ×14 = 24 ModY = a + c ⋅ Mod X = 80 − 4 ×15 = 20 σ Y =| c | ⋅σ X = 4 × 3.6 = 14.4 σ 14.4 C.VY = Y = = 45% µY 32 3( µY − MedY ) 3(32 − 24) SkY = = = 1.66667 σY 14.4...
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