Q1= 2.2110.6221.2331.6441.9551.5662.1772.3812.3922.510 3 2.81142.9>6#*.,6(/? (*:,6#2@(+86(,-#*.&/"+The first quartile, Q1, is the median of values below M.The third quartile, Q3, is the median of values above M.M = 3.4Q3= 4.351142.912 5 3.313 3.414 1 3.615 2 3.716 3 3.817 4 3.918 5 4.119 6 4.220 7 4.521 1 4.722 2 4.923 3 5.324 4 5.625 5 6.1of values below M.values above M.!"#$:56(3#53.5#+%/&@(86%'8+*Problem: Heights, x, is N(64.5, 2.5).For what proportion of individuals is x < 67?Solution:Ask: How far is c = 67 from µ = 64.5 in units of σ = 2.5?(c – µ) / σ = (67 – 64.5) / 2.5 = 1Translate: z = (x – µ) / σ is N(0, 1)For what proportion of individuals is z < 1?Calculate: normsdist(1) = 0.84!"#$:56(3#53.5#+%/&@(86%'8+*(N3/&+O68-95-99.7 rule: Proportion with -1 < z < 1 is 0.68Equally divide remaining between z < -1 and z > 1Proportion with z < 1 is 0.16 + 0.68 = 0.840.680.160.16P#53.5#+%/&(/? (Q#,6#(-6+766&RProblem: Proportion with c1< z < c2Solution: (prop. with z < c2) – (prop. with z < c1)Example: Proportion with 1.4 < z < 2.2.normsdist(2.2) –normsdist(1.4)normsdist(2.2) –normsdist(1.4)= 0.9861 – 0.9192 = 0.0669S#3T7#,2(3#53.5#+%/&*Problem: For what c is p the proportion with z < c?Solution: c = normsinv(p)Examples:normsinv(0.84) = 1 0.680.160.16normsinv(0.84) = 1 normsinv(0.16) = -1Problem: MPG, x, of compact cars is N(25.7, 5.88).For what c does 10% of compact cars have x > c?Solution: First, normsinv(0.90) = 1.28Translate: z = (x – µ) / σ is N(0, 1)!"#$:56(3#53.5#+%/&@($:'10% of compact carshave z > 1.28 = (c – µ) / σSolve: 1.28 = (c –24.7) / 5.88 ! c = 24.7 + (1.28)(5.88) = 33.25. A study looks at the relationship between employee salaryand hours worked at a certain company. The average yearlyemployee salary is $60,000 and has a standard deviation of$5,000. The average number of hours worked per week is42 with a standard deviation of 7 hours. We know that 49%of variation in salary is explained by the variation in hoursworked. If an employee works 50 hours per week, what doyou predict his/her salary to be? (Assume that as salary in-creases, so does hours wor ked.)From the text:¯y = 60, 000, sy= 5, 000¯x = 42, sx= 7r2= 0.49, positive relationship, ⇒ r = +√0.49 = +0.7Regression line:b1= rsysx= (0.7)50007= 500b0= ¯y − b1¯x = 60, 000 − (500)(42) = 39, 000⇒ˆy = b0+ b1x = 36, 500 + 500xPrediction: Plug in x = 50⇒ˆy = b0+ b1(50) = 39, 000 + (500)(50) = 64, 000DE*3602F,2G%*004,0'>204,#%+&#32/A couple wants to have three children. Observe the possible sequences of boys (B) and girls (G).S = { BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG }BBB - BBBG - BBGAssign equal probability of 1/8 to each outcomeBGB - BGBG - BGGA = “exactly two girls” = { BGG, GBG, GGB }BB - GBBG - GBGP(A) = P(BGG) + P(GBG) + P(GGB)=1/8+1/8+1/8GB - GGBG-GGGG 1/8 1/8 1/8 = 3/8G -GGGDE*3602F,92(1#"$H/ @*.Empirical probabilities of “first digits” in financial docs1stdigit123456789Probability0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.0460200.250.300.35bilityProbability histogram0.000.050.100.150.20ProbabP(1stdigit t6) = 0.067 + 0.058 + 0.051 + 0.046 = 0.222123456789Outcomes(dgt 6) 0 06 0 058 0 05 0 0 6 0DE*3602F,-.#,$'2,"#00/Thirty-six possible die rolls, equal probabilities:P(sum is 5)=4/36=0.111P(sum is 5) 4/36 0.111P(doubles) = 6/36 = 0.167, etc.Note: X = “sum” is an example of a “random variable” (more later)DE*3602F,I('1#"3,$2(/'+4,&%"72Probabilities of a random number generator, S={numbersbetween0and1}S { numbers between 0 and 1 }P(0 3dXd07)=07–03=04P(0.3 dXd0.7) = 0.7 –0.3 = 0.4DE*3602F,J2(2"*0,%('1#"3,$2(/'+4,&%"72Probabilities of a custom random number generator, S={numbersbetweenc1andc2}S { numbers between c1and c2}P(a d X d b) = (b – a) / (c2– c1)DE*3602F,5%3,#1,+.#,"*($#3,(%382"/Sum of two numbers from a random number generator, S={numbersbetween0and2}S { numbers between 0 and 2}heightheightP(X>13)=!bh= ! (2–13)(2–13)=0245baseP(X> 1.3) = !bh= ! (2 –1.3) (2 –1.3) = 0.245DE*3602F,K#"3*0,&%"72/X = ACT college entrance exam scoresSupposeXis N(P=18 6V=59)Suppose Xis N(P18.6, V 5.9)Probability interpretation: X is the score of a randomly selected studentselected student=*($#3,7*"'*802/A random variable, X, is an idealization of quantitative data recorded from many repetitions of a chance-yphappening.BB - BBBG-BBGExample: A couple wants to have three childrenBGG BBGB - BGBX = # girlsBB - GBBG - BGGS = { 0, 1, 2, 3 }BGG - GBGB - GGBGG - GGG!"#8*8'0'+4,$'/+"'8%+'#(,#1,*,"*($#3,7*"'*802The probability distribution of a random variable is its assignment of probabilities in an underlying probability gp ygpymodel.Example: X= # girls among three childrenP(X=0)=P(BBB)=1/8P(X= 0) = P(BBB) = 1/8P(X = 1) = P(BBG) + P(BGB) + P(GBB) = 3/8P(X=2)=P(BGG)+P(GBG)+P(GGB)=3/8P(X 2) P(BGG) P(GBG) P(GGB) 3/8P(X = 3) = P(BBB) = 1/8:;%-.'6< =0/&"'20%-#*/0(8"660>8&'+"6*A couple wants to have three children. X = # girls.xi0123pi1/8 3/8 3/8 1/8PX= ¦ xipi= (0)(1/8) + (1)(3/8) + (2)(3/8) + (3)(1/8) = 1.5 Note X cannot take the value PX= 1.5. Interpret “expected” # girls as X = 1 or X = 2 with no leaning either wayy:;%-.'6< =0/&"'20%-#*/0(8"660>8&'+"6*A couple wants to have three children. X = # girls.xi0123pi1/8 3/8 3/8 1/8PX= ¦ xipi= (0)(1/8) + (1)(3/8) + (2)(3/8) + (3)(1/8) = 1.5VX2= ¦ (xi– PX)2pi= (-1.5)2(1/8) + (-0.5)2(3/8) + (0.5)2(3/8) + (1.5)2(1/8) = 0.75VX= VX2= 0.75 = 0.87:;%-.'6< B*562(-6*(2Setup: rates of returnsX=T-billsX TbillsY = A certain index fundZ = my portfolio = 0.2 X+ 0.8 YypInterpret 1950-2003 history:PX= 5.0%, VX= 2.9%PY= 13.2%, VY= 17.6%011UXY= -0.11:;%-.'6< B*562(-6*(20@>#*(&*36+ASummary of my portfolio: Z = 0.2 X + 0.8 YPZ= 0.2 PX+ 0.8 PY= (0.2)(5.0) + (0.8)(13.2) = 11.56%V0.2X2= (0.2)2VX2= (0.2)2(2.9)2= 0.340.2X()X()()V0.8Y2= (0.8)2VY2= (0.8)2(17.6)2= 198.25VZ2=V02X2+V08Y2+2UXYV02XV08YVZ V0.2X+ V0.8Y+ 2 UXYV0.2XV0.8Y= 0.34 + 198.25 + 2 (-0.11) (0.34) (198.25)= 196.79Interpretvia689599 7 ruleVZ= VZ2= 196.79 = 14.03%Interpret via68-95-99.7 rule(assuming a Normal distribution):;%-.'6< I#(('&*/0#.6"%(&#*Fill-level of bottling machine has P = 298 and V = 3millilitersmilliliters What is the probability that the fill-level of some dl ltdbttlil th295l?randomly selected bottle is less than 295ml?I don’t know:;%-.'6< I#(('&*/0#.6"%(&#*0@>#*(&*36+AFill-level of bottling machine has P = 298 and V = 3millilitersmilliliters What is the probability that the average