MATH 4426 HW7 solutions. April 15, Recall, Z is said to have a standard normal distribution (denoted Z N(0, 1)) if its pdf is

Transcription

1 MATH 446 HW7 solutions April 5, 5 Recall, Z is said to have a standard normal distribution (denoted Z N(, )) if its pdf is f Z (z) = p e z /,z (, ). Table A. (pp ) tabulates values of its cdf (denoted.) Examples: P (Z apple.63) = (.63) =.9484; P (Z >.56) = (.56) =.877 =.73; P (.56 <Zapple.63) = (.63) (.56) = = Find the following: a. P (.34 apple Z<.8) =.8) (.34) = =.738. b. P (Z >.) = P (Z apple.) = (.) =.9778 =.. c. The (approximate) z such that P (Z > z) =. () (z) =.98 () z =.5 (approximately).. Next, X N(µ, )ifitspdfis f X (x) = p e (x µ) /,x (, ). Note the standard normal is a special case: µ =, =. Prove the following: X N(µ, ) =) X µ N(, ). (We abbreviate this by writing Z = X µ.) Prove this by letting, say, W =(X µ)/ and deriving the pdf of W ;asusual,it seasiestfirsttofindthecdfofw as an integral, then di erentiate via the fundamental theorem of calculus, part. For w (, ), F W (w) =P (W apple w) =P ( X µ apple w) =P (X apple µ + w) =

2 Z µ+ w p e (x µ) / dx, so f W (w) = d dw F W (w) = Z d µ+ w p e (x µ) / dx = p e (µ+ w µ) / d (µ + w) = dw dw which is the pdf of a N(, ) rv. p e ( ) /( ) = p e w /, 3. Problems and now allow you to calculate probabilities for any normal random variable by, if necessary, first transforming it to an equivalent problem with the standard normal. Suppose the weight-bearing capacity of a cable is 7 kip. (kip is a unit equivalent to pounds). Suppose we need to lift individual weights at di erent times. If the weight to be lifted (the load e ect ) is normally distributed with mean 5 kip and standard deviation kip, what s the probability the cable will not break? Let X be the weight, then X N(5, 44). The probability the cable will not break is P (X <7) = P ( X 5 < 7 5 = P (Z <.67) = (.67) = Suppose you are designing a cable to handle load e ects that are normally distributed with mean 5 kip and standard deviation kip and you want to be certain that there s only a % probability that a load from this distribution will exceed the capacity. What should the capacity be? Again, let X be the weight, then X N(5, 44). We wish to find a value x so that P (X x) =.. We transform the problem to the standard normal:. = P (X x) =P ( X 5 x 5 )=P(Z x 5 ) From problem (c), P (Z >z)=. when z =.5, so we must have x 5 x =5+(.5) = =.5, or 5. Now suppose the weight-bearing capacity of the cable, denoted R, isanormallydis- tributed random variable with mean kip and standard deviation 8 kip, while the load e ects, denoted S, arenormallydistributedwithmean5kipandstandarddeviationkip. Suppose that R and S are independent. Find the probability the cable will break under a randomly chosen load. (Hint: P (R <S)=P (R S<.)

3 R N(, 34), S N(5, 44) independent =) R S N( 5, 34+( ) 44) = N(7, 468). Hence P (R S<) = P ( R S 7 p 468 < 7 p 468 = P (Z < 3.3) = ( 3.3) = section 3.9, 6 Let X= thetotalgainandx i =thegainondayi, fori =,...,n.then()x = P n i= X i and () E(X i )=(/8)p +( /8)q =(p q)/8. Hence E(X) =E X i = i= E(X i )= i= (p q)/8 =n(p q)/8. i= 7. section 3.9, 8 Let X be the number rolled on the first die and Y on the second. Then E(X) =E(Y )= (/6)+(/6)+3(/6)+4(/6)+5(/6)+6(/6) = 7/. By independence of X, Y, E(XY )= E(X)E(Y )=(7/) =49/4. 8. section 3.9, X U(, ), Y U(, ) =) E(Y )=E(X )= Therefore E(X + Y )=E(X )+E(Y )=/3. Z x dx = x 3 /3 =/3. 9. section 3.9, Recall the mean for a Binomial(k,p) is kp and the variance is kpq, whereq = since X, Y are independent, p. Hence, E(X) =np X,V(X) =np X q x,e(y )=mp Y,V(Y )=mp Y q y =) E(W )=E(4X +6Y )=4E(X)+6E(Y )=4np X +6mp Y 3

4 and V (W )=V (4X +6Y )=6E(X)+36E(Y )=6np X q X +36mp Y q y. Note the result for E(W )doesnotdependontheindependenceofx, Y but the result for the variance does.. section 3., 6 Y,Y,...,Y n arandomsamplefromf Y (y) =e y,y. We need to find the pdf for Y min. The common cdf is F Y (y) = R y e t dt = e y,fory. Thus Y min(y) has pdf f Ymin (y) =n( ( e y )) n e y = n(e y ) n = ne ny,y. Next, we have Therefore, P (Y min <.) = Z. ne ny dy = e ny. = e.n. P (Y min <.) >.9 () e.n >.9 () e.n <. ().n <ln(.) () n>ln(.)/(.) =.5 =) n.. section 3.. (You re looking for the expected value of what order statistic?) We have lifetimes Y,...,Y n independent with common pdf f Y (y) = e y,y. It s easily checked that the common cdf is F Y (y) = e y,y, so that Y min has pdf n( ( e y ) n e y = n e n y.we reaskedtofind E(Y min )= Z yn e n y = (integration by parts) ye n y + Z e n y dy. The first expression is (after an application of L Hôpital s rule) and the second is n e n y =+ n = n.. Three sealed bid amounts are independent random variables with common density function f(x) =6x 7, for x>, where x is the amount of a claim in thousands of dollars. a. Find the pdf of the largest bid. The common cdf is F (x) = Z x 6t 7 dt = t 6 x = x 6,x>. 4

5 Therefore X max has pdf f Xmax (x) =3( x 6 ) 6x 7 =8( x 6 ) x 7,x>. b. Find the probability the maximum bid is greater than. We integrate the pdf of X max over [, ). We could do a substitution or multiply it out: P (X max > ) = Z 8( x 6 ) x 7 dx =8 Z x 7 x 3 + x 9 dx = 8 6x + =3(/) 6 3(/) +(/) 8 = x 8x 8 c. Find the expected value of the largest bid. 8 Z E(X max )= Z x8( x 6 ) x 7 dx = x 6 x + x 8 dx =8( 5x + 5 x + 7x ) 7 8( )=.386. = 5