Exam 5: Joint Probability Distributions and Random Samples

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Let X1,X2, and X3X _ { 1 } , X _ { 2 } \text {, and } X _ { 3 } represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent normal random variables with expected values μ1,μ1, and μ3 and variances σ12,σ12, and σ32\mu _ { 1 } , \mu _ { 1 } \text {, and } \mu _ { 3 } \text { and variances } \sigma _ { 1 } ^ { 2 } , \sigma _ { 1 } ^ { 2 } \text {, and } \sigma _ { 3 } ^ { 2 } \text {, } respectively. a. If μ=μ2=μ3=65 and σ12=σ22=σ32=20\mu = \mu _ { 2 } = \mu _ { 3 } = 65 \text { and } \sigma _ { 1 } ^ { 2 } = \sigma_ { 2 } ^ { 2 } = \sigma_ { 3 } ^ { 2 } = 20 \text {, } Calculate P(X1+X2+X3210)P \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \leq 210 \right) What is P(150X1+X2+X3210)?P \left( 150 \leq X _ { 1 } + X _ { 2 } + X _ { 3 } \leq 210 \right) ? b. Using the μ2s and σ2s\mu _ { 2 } ^ { \prime } s \text { and } \sigma _ { 2 } ^ { \prime } s given in part (a), calculate P(Xˉ59) and P(62Xˉ68)P ( \bar { X } \geq 59 ) \text { and } P ( 62 \leq \bar { X } \leq 68 ) c. Using the μ2s and σ2s\mu _ { 2 } ^ { \prime } s \text { and } \sigma _ { 2 } ^ { \prime } s given in part (a), calculate P(10X1.5X2.5X35)P \left( - 10 \leq X _ { 1 } - .5 \mathrm { X } _ { 2 } - .5 X _ { 3 } \leq 5 \right) d. If μ1=40,μ2=50,μ3=60,σ12=10,σ22=12, and σ32=14\mu _ { 1 } = 40 , \mu _ { 2 } = 50 , \mu _ { 3 } = 60 , \sigma _ { 1 } ^ { 2 } = 10 , \sigma _ { 2 } ^ { 2 } = 12 \text {, and } { \sigma_ { 3 } ^ { 2 } }= 14 \text {, } calculate P(X1+X2+X3160) and P(X1+X22X3)P \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \leq 160 \right) \text { and } P \left( X _ { 1 } + X _ { 2 } \geq 2 X _ { 3 } \right)

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a. E(X1+X2+X3)=195,V(X1+X1+X3)=60,σx1+x2+x3=7.746E \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \right) = 195 , V \left( X _ { 1 } + X _ { 1 } + X _ { 3 } \right) = 60 , \sigma _ { x _ { 1 } + x _ { 2} + x _ { 3 } } = 7.746 P(X1+X2+X3210)=P(Z2101957.746)=P(Z1.94)=.9738P \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \leq 210 \right) = P \left( Z \leq \frac { 210 - 195 } { 7.746 } \right) = P ( Z \leq 1.94 ) = .9738 P(175X1+X2+X3210)=P(2.58Z1.94)=.9689P \left( 175 \leq X _ { 1 } + X _ { 2 } + X _ { 3 } \leq 210 \right) = P ( - 2.58 \leq Z \leq 1.94 ) = .9689 b. μzˉ=μ=65,σzˉ=σzˉn=123=2.582\mu _ { \bar { z } } = \mu = 65 , \sigma _ { \bar { z } } = \frac { \sigma _ { \bar { z } } } { \sqrt { n } } = \frac { \sqrt { 12 } } { \sqrt { 3 } } = 2.582 P(Xˉ59)=P(Z59652.582)=P(Z2.232)=.9898P (\bar X \geq 59 ) = P \left( Z \geq \frac { 59 - 65 } { 2.582 } \right) = P ( Z \geq - 2.232 ) = .9898 P(62Xˉ68)=P(1.16Z1.16)=.754P ( 62 \leq \bar { X } \leq 68 ) = P ( - 1.16 \leq Z \leq 1.16 ) = .754 c. E(X1.5X25X3)=0E \left( X _ { 1 } - .5 X _ { 2 } - 5 X _ { 3 } \right) = 0 V(X1.5X2.5X3)=σ12+.25σ22+.25σ32=30,sd=5.4772V \left( X _ { 1 } - .5 X _ { 2 } - .5 X _ { 3 } \right) = \sigma _ { 1 } ^ { 2 } + .25 \sigma _ { 2 } ^ { 2 } + .25 \sigma _ { 3 } ^ { 2 } = 30,s d = 5.4772 P(10X1.5X2.5X35)=P(1005.4772Z505.4772)P \left( - 10 \leq X _ { 1 } - .5 X _ { 2 } - .5 X _ { 3 } \leq 5 \right) = P \left( \frac { - 10 - 0 } { 5.4772 } \leq Z \leq \frac { 5 - 0 } { 5.4772 } \right) =P(1.83Z.91)=.8186.0336=.785= P ( - 1.83 \leq Z \leq .91 ) = .8186 - .0336 = .785 d. E(X1+X2+X3)=150,V(X1+X1+X3)=36,σx1+x2+x3=6E \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \right) = 150 , V \left( X _ { 1 } + X _ { 1 } + X _ { 3 } \right) = 36 , \sigma _ { x _ { 1 } + x _ {2 } + x _ { 3 } } = 6 P(X1+X2+X3200)=P(Z1601506)=P(Z1.67)=.9525P \left( X _ { 1 } + X _ { 2 } + X _ { 3 } \leq 200 \right) = P \left( Z \leq \frac { 160 - 150 } { 6 } \right) = P ( Z \leq 1.67 ) = .9525 We want P(X1+X22X3), or written another way, P(X1+X22X30)P \left( X _ { 1 } + X _ { 2 } \geq 2 X _ { 3 } \right) \text {, or written another way, } P \left( X _ { 1 } + \mathrm { X } _ { 2 } - 2 X _ { 3 } \geq 0 \right) \text {. } E(X1+X22X3)=40+502(60)=30E \left( X _ { 1 } + X _ { 2 } - 2 X _ { 3 } \right) = 40 + 50 - 2 ( 60 ) = - 30 V(X1+X22X3)=σ12+σ22+4σ32=78,36,sd2=8.832,soV \left( X _ { 1 } + X _ { 2 } - 2 X _ { 3 } \right) = σ _ { 1 } ^ { 2 } + σ_ { 2 } ^ { 2 } + 4 σ _ { 3 } ^ { 2 } = 78,36,s d ^ { 2 } = 8.832 , \mathrm { so }
P(X1+X22X30)=P(Z0(30)8.832)=P(Z3.40)=.0003P \left( X _ { 1 } + X _ { 2 } - 2 X _ { 3 } \geq 0 \right) = P \left( Z \geq \frac { 0 - ( - 30 ) } { 8.832 } \right) = P ( Z \geq 3.40 ) = .0003

Let X denote the number of brand X VCRs sold during a particular week by a certain store. The pmf of X is x 0 1 2 3 4 .1 .2 .3 .25 .15 (x) Seventy percent of all customers who purchase brand X VCRs also buy an extended warranty. Let Y denote the number of purchasers during this week who buy an extended warranty. a. What is P(X = 4, Y = 2)? [Hint: This probability equals P(Y = 2/X = 4)  . \text { . } P(X = 4); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.] b. Calculate P(X =Y). c. Determine the joint pmf of X and Y and then the marginal pmf of Y.

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a. p(4,2)=P(Y=2X=4)P(X=4)=[(42)(.7)2(.4)2](.15)=.0397p ( 4,2 ) = P ( Y = 2 \mid X = 4 ) \cdot P ( X = 4 ) = \left[ \left( \begin{array} { l } 4 \\2\end{array} \right) ( .7 ) ^ { 2 } ( .4 ) ^ { 2 } \right] \cdot ( .15 ) = .0397 b. P(X=Y)=p(0,0)+p(1,1)+p(2,2)+p(3,3)+p(4,4)=.1+(.2)(.7)+(.3)(.7)2+P ( X = Y ) = p ( 0,0 ) + p ( 1,1 ) + p ( 2,2 ) + p ( 3,3 ) + p ( 4,4 ) = .1 + ( .2 ) ( .7 ) + ( .3 ) ( .7 ) ^ { 2 } + (.25)(.7)3+(.15)(.7)4=.5088( .25 ) ( .7 ) ^ { 3 } + ( .15 ) ( .7 ) ^ { 4 } = .5088 c. p(x,y)=0 unless y=0,1,,x;x=0,1,2,3,4p ( x , y ) = 0 \text { unless } y = 0,1 , \ldots , x ; x = 0,1,2,3,4 For any such pair, p(x,y)=P(Y=yX=x)P(X=x)=(xy)(7)(.3)xypx(x)p ( x , y ) = P ( Y = y \mid X = x ) \cdot P ( X = x ) = \left( \begin{array} { l } x \\y\end{array} \right) ( 7 ) ( .3 ) ^ { x -y} \cdot p _ { x } ( x ) py(4)=p(y=4)=p(x=4,y=4)=p(4,4)=(.7)4(.15)=.0360p_{y} ( 4 ) = p ( y = 4 ) = p ( x = 4 , y = 4 ) = p ( 4,4 ) = ( .7 ) ^ { 4 } \cdot ( .15 ) = .0360 py(3)=p(3,3)+p(4,3)=(.7)3(.25)+(43)(.7)3(.3)(.15)=.1475p_{y} ( 3 ) = p ( 3,3 ) + p ( 4,3 ) = ( .7 ) ^ { 3 } ( .25 ) + \left( \begin{array} { l } 4 \\3\end{array} \right) ( .7 ) ^ { 3 } ( .3 ) ( .15 ) = .1475 py(2)=p(2,2)+p(3,2)+p(4,2)=(.7)2(.3)+(32)(.7)2(.3)(.25)p _{y} ( 2 ) = p ( 2,2 ) + p ( 3,2 ) + p ( 4,2 ) = ( .7 ) ^ { 2 } ( .3 ) + \left( \begin{array} { l } 3 \\2\end{array} \right) ( .7 ) ^ { 2 } ( .3 ) ( .25 ) +(42)(.7)2(.3)2(.15)=.2969+ \left( \begin{array} { l } 4 \\2\end{array} \right) ( .7 ) ^ { 2 } ( .3 ) ^ { 2 } ( .15 ) = .2969 Py(1)=p(1,1)+p(2,1)+p(3,1)+p(4,1)=(.7)(.2)+(21)(.7)(.3)(.3)P_{y} ( 1 ) = p ( 1,1 ) + p ( 2,1 ) + p ( 3,1 ) + p ( 4,1 ) = ( .7 ) ( .2 ) + \left( \begin{array} { l } 2 \\1\end{array} \right) ( .7 ) ( .3 ) ( .3 ) (31)(.7)(.3)2(.25)+(41)(.7)(.3)3(.15)=.3246\left( \begin{array} { l } 3 \\1\end{array} \right) ( .7 ) ( .3 ) ^ { 2 } ( .25 ) + \left( \begin{array} { l } 4 \\1\end{array} \right) ( .7 ) ( .3 ) ^ { 3 } ( .15 ) = .3246 p,(0)=1[.3246+.2969+.1475+.0360]=.1950p , ( 0 ) = 1 - [ .3246 + .2969 + .1475 + .0360 ] = .1950

Abby and Bianca have agreed to meet for lunch between noon and 1:00 P.M. Denote Abby's arrival time by X, Bianca's by Y, and suppose X and Y are independent with pdf's. fx(x)={3x20x10 otherwise f _ { x } ( x ) = \left\{ \begin{array} { l l } 3 x ^ { 2 } & 0 \leq x \leq 1 \\0 & \text { otherwise }\end{array} \right.  Abby and Bianca have agreed to meet for lunch between noon and 1:00 P.M. Denote Abby's arrival time by X, Bianca's by Y, and suppose X and Y are independent with pdf's.  f _ { x } ( x ) = \left\{ \begin{array} { l l }  3 x ^ { 2 } & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array} \right.     f _ { y } ( y ) = \left\{ \begin{array} { l l }  2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array} \right.  What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: h(X, Y ) = |X - Y|.] fy(y)={2y0y10 otherwise f _ { y } ( y ) = \left\{ \begin{array} { l l } 2 y & 0 \leq y \leq 1 \\0 & \text { otherwise }\end{array} \right. What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: h(X, Y ) = |X - Y|.]

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E[h(X,Y)]=0101xy6x2ydxdy=2010x(xy)6x2ydydxE [ h ( X , Y ) ] = \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } | x - y | \cdot 6 x ^ { 2 } y d x d y = 2 \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { x } ( x - y ) \cdot 6 x ^ { 2 } y d y d x = 12010x(x3yx2y2)dydx=1201x56dx=1312 \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { x } \left( x ^ { 3 } y - x ^ { 2 } y ^ { 2 } \right) d y d x = 12 \int _ { 0 } ^ { 1 } \frac { x ^ { 5 } } { 6 } d x = \frac { 1 } { 3 }

In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let X = the number of trees planted in sandy soil that survive 1 year and Y = the number of trees planted in clay soil that survive 1 year. If the probability that a tree planted in sandy soil will survive 1 year is .7 and the probability of 1-year survival in clay soil is .6, compute an approximation to P(6XY6)P ( - 6 \leq X - Y \leq 6 ) (do not bother with the continuity correction).

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Two components of a minicomputer have the following joint pdf for their useful lifetimes X and Y: f(x,y)={xex(1+y)x0 and y00 otherwise f ( x , y ) = \left\{ \begin{array} { l l } x e ^ { - x ( 1 + y ) } & x \geq 0 \text { and } y \geq 0 \\0 & \text { otherwise }\end{array} \right. a. What is the probability that the lifetime X of the first component exceeds 3? b. What are the marginal pdf"s of X and Y? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds 3?

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The joint pdf of pressures for right (X) and left (Y) front tires is given by f(x,y)={K(x2+y2)20x30,20y300 otherwise f ( x , y ) = \left\{ \begin{array} { c l } K \left( x ^ { 2 } + y ^ { 2 } \right) & 20 \leq x \leq 30,20 \leq y \leq 30 \\0 & \text { otherwise }\end{array} \right. . a. Determine the conditional pdf of Y given that X = x and the conditional pdf of X given that Y = y if you are given FX(x)=10kx2+.05 for 20x30 and FY(y)=10 ky2+.05 for 20y30F _ { X } ( x ) = 10 \mathrm { kx } ^ { 2 } + .05 \text { for } 20 \leq x \leq 30 \text { and } F _ { Y } ( y ) = 10 \mathrm {~ky} ^ { 2 } + .05 \text { for } 20 \leq y \leq 30 b. If the pressure in the right tire is found to be 22 psi, what is the probability that the left tire has a pressure of at least 25 psi? Compare this to P(Y25)P ( Y \geq 25 ) c. If the pressure in the right tire is found to be 22 psi, what is the expected pressure in the left tire, and what is the standard deviation of pressure in this tire?

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Suppose your waiting time for a bus in the morning is uniformly distributed on [0,5], whereas waiting time in the evening is uniformly distributed on [0,10] independent of morning waiting time. a. If you take the bus each morning and evening for a week, what is your total expected waiting time? [Hint: Define random variables X1,X10X _ { 1 } , \ldots X _ { 10 } and use a rule of expected value.) b. What is the variance of your total waiting time? c. What are the expected value and variance of the difference between morning and evening waiting times on a given day? d. What are the expected value and variance of the difference between morning waiting time and total evening waiting time for a particular week?

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Show that if X and Y are independent random variables, then E(XY)=E(X)E(Y)E ( X Y ) = E ( X ) \cdot E ( Y )

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Let X be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of X is as follows: x 1 2 3 4 p(x) .4 .3 .2 .1 a. Consider a random sample of size n = 2 (two customers), and let Xˉ\bar { X } be the sample mean number of packages shipped. Obtain the probability distribution of Xˉ\bar { X } . b. Refer to part (a) and calculate P(Xˉ2.5)P ( \bar { X } \leq 2.5 ) c. Again consider a random sample of size n = 2, but now focus on the statistic R = the sample range (difference between the largest and smallest values in the sample). Obtain the distribution of R. [Hint: Calculate the value of R for each outcome and use the probabilities from part (a).] d. If a random sample of size n = 4 is selected, what is P(Xˉ1.5)P ( \bar { X } \leq 1.5 ) ? (Hint: You should not have to list all possible outcomes, only those for which xˉ1.5.)\bar { x } \leq 1.5 . )

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The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi. a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9950 and 10,250? b. If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information?

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Show that if Y=aX+b(a0), then Corr(X,Y)=+1 or 1Y = a X + b ( a \neq 0 ) \text {, then } \operatorname { Corr } ( X , Y ) = + 1 \text { or } - 1 Under what conditions will ρ=+1?\rho = + 1 ?

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Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable-X for the right tire and Y for the left tire, with joint pdf fk(x,y)={K(x2+y2)20x30,20y300 otherwise f k ( x , y ) = \left\{ \begin{array} { l l } K \left( x ^ { 2 } + y ^ { 2 } \right) & 20 \leq x \leq 30,20 \leq y \leq 30 \\0 & \text { otherwise }\end{array} \right. a. What is the value of K? b. What is the probability that both tires are underfilled? c. What is the probability that the difference in air pressure between the two tires is at most 2 psi? d. Determine the (marginal) distribution of air pressure in the right tire alone. e. Are X and Y independent random variables?

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Three different roads feed into a particular freeway entrance. Suppose that during a fixed time period, the number of cars coming from each road onto the freeway is a random variable, with expected value and standard deviation as given in the table. Road 1 Road 2 Road 3 Expected value 750 1000 550 16 24 18 Standard deviation a. What is the expected total number of cars entering the freeway at this point during the period? (Hint: Let Xi,= the number from road i.) X _ { i } , = \text { the number from road } i \text {.) } b. What is the variance of the total number of entering cars? Have you made any assumptions about the relationship between the numbers of cars on the different roads? c. With Xˉi\bar{X} _ { i} denoting the number of cars entering from road I during the period, suppose that Cov(X1,X2)=80,Cov(X1,X3)=90, and Cov(X2,X3)=100\operatorname { Cov } \left( X _ { 1 } , \quad X _ { 2 } \right) = 80 , \quad \operatorname { Cov } \left( X _ { 1 } , \quad X _ { 3 } \right) = 90 \text {, and } \operatorname { Cov } \left( X _ { 2 } , X _ { 3 } \right) = 100 (so that the three streams of traffic are not independent). Compute the expected total number of entering cars and the standard deviation of the total.

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It is known that 80% of all brand A zip drives work in a satisfactory manner throughout the warranty period (are "success"). Suppose that n = 10 drives are randomly selected. Let X = the number of successes in the sample. The statistic X/n is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value of X/n is .3, corresponding to X = 3. What is the probability of this value (what kind of random variable is X)?]

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The lifetime of a certain type of battery is normally distributed with mean value 12 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

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A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries (1000's of dollars) is as follows: 1 1 2 2 3 3 Office Employee 1 2 3 4 5 6 Salary 19.7 23.6 20.2 23.6 15.8 19.7 a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary Xˉ\bar { X } b. Suppose one of the three offices is randomly selected. Let X1 and X2X _ { 1 } \text { and } X _ { 2 } denote the salaries of the two employees. Determine the sampling distribution of Xˉ\bar { X } c. How does E(Xˉ)E ( \bar { X } ) from parts (a) and (b) compare to the population mean salary μ?\mu ?

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A particular brand of dishwasher soap is sold in three sizes: 25oz, 40oz, and 65 oz. Twenty percent of all purchasers select a 25 oz box, fifty percent select a 40 oz box, and the remaining thirty percent choose a 65 oz box. Let X1 and X2X _ { 1 } \text { and } X _ { 2 } denote the package sizes selected by two independently selected purchasers. a. Determine the sampling distribution of Xˉ\bar { X } , calculate E(Xˉ)E ( \bar { X } ) , and compare to μ.\mu . b. Determine the sampling distribution of the sample variance S2, calculate E(S2), and compare to σ2S ^ { 2 } \text {, calculate } E \left( S ^ { 2 } \right) \text {, and compare to } \sigma ^ { 2 } \text {. }

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An instructor has given a short test consisting of two parts. For a randomly selected student, let X = the number of points earned on the first part and Y = the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table. p(x,y) 0 5 10 15 0 .02 .06 .02 .10 5 .04 .15 .20 .10 10 .01 .15 .14 .01 a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score E(X + Y)? b. If the maximum of the two scores is recorded, what is the expected recorded score?

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The number of parking tickets issued in Grand Rapids on any given weekday has a Poisson distribution with parameter λ^=50\hat\lambda = 50 What is the approximate probability that a. Between 40 and 70 tickets are given out on a particular day? (Hint: When λ^\hat\lambda is large, a Poisson random variable has approximately a normal distribution.) b. The total number of tickets given out during a 5-day week is between 215 and 265?

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