STAT 516 Answers Homework 5 March 3, 2008 Solutions by Mark Daniel Ward PROBLEMS

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1 STAT 5 Answers Homework 5 March 3, 28 Solutions by Mark Daniel Ward PROBLEMS We always let Z denote a standard normal random variable in the computations below. We simply use the chart from page 222 of the Ross book, when working with a standard normal random variable; of course, a more accurate computation is possible using the computer. 2. We first compute = = 2C Cxe x/2 dx = C /2 (x e x/2 x= e x/2 dx = 4Ce x/2 x= = 4C e x/2 /2 dx Thus C = /4. Now we compute the probability that the system functions for at least 5 months: P (X 5 = 5 4 xe x/2 dx = (x e x/2 e x/2 4 /2 x=5 5 /2 dx = ( e 5/2 + 2 e x/2 dx = ( e 5/2 4 e x/2 = x=5 2 e 5/2 3a. The function f(x = 5 C(2x x 3 if < x < 5/2 otherwise cannot be a probability density function. To see this, note 2x x 3 = x(x 2(x + 2. If C =, then f(x = for all x, so f(x =, but every probability density function has f(x =. If C >, then f(x < on the range x (, 2, but every probability density function has f(x for all x. If C <, then f(x < on the range x ( 2, 5/2, but every probability density function has f(x for all x. So no value of C will satisfy all of the properties needed for a probability density function. 3b. The function C(2x x 2 if < x < 5/2 f(x = otherwise cannot be a probability density function. To see this, note 2x x 2 = x(x 2. If C =, then f(x = for all x, so f(x =, but every probability density function has f(x =. If C >, then f(x < on the range x (, 2, but every probability density function has f(x for all x. If C <, then f(x < on the range x (2, 5/2, but every probability density function has f(x for all x. So no value of C will satisfy all of the properties needed for a probability density function.

2 2 5. Let X denote the volume of sales in a week, given in thousands of gallons. We have the density of X, and we want a with P (X > a =.. Note that we need < a <. We have. = P (X > a = f(xdx = 5( x 4 ( x5 dx+ dx = 5 ( 5 = ( a 5 a So. = ( a 5, and thus 5. = a, so a = a x=a a. We compute ( E[X] = xf(x dx = (x( dx + (x (x(e x/2 dx = x 2 e x/2 dx 4 4 = (x 2 e x/2 4 /2 2x e x/2 x= /2 dx = (x 2 e x/2 4 /2 + 4 xe x/2 dx x= = (x 2 e x/2 4 /2 + 4x e x/2 e x/2 x= /2 4 x= /2 dx = (x 2 e x/2 4 /2 + 4x e x/2 /2 + 8 e x/2 /2 = 4 ( = 4 b. We compute = f(x dx = x= dx + and thus c = 3/4. Now we compute E[X] = c. We compute xf(x dx = ( x 2 = (3/4 2 x4 4 E[X] = xf(x dx = x= x= c( x 2 dx + (x( dx + 5 = (3/4 (x( dx + x= ( 4 4 dx = c x(3/4( x 2 dx + 5 = (x x3 = c(4/3 3 x= (x( dx x 5 x 2 dx = 5 ln x x=5 = + a. The times in which the passenger would go to destination A are 7:5 7:5, 7:2 7:3, 7:35 7:45, 7:5 8:, i.e., a total of 4 out of minutes. So 2/3 of the time, the passenger goes to destination A. b. The times in which the passenger would go to destination A are 7: 7:5, 7:2 7:3, 7:35 7:45, 7:5 8:, 8:5 8:, i.e., a total of 4 out of minutes. So 2/3 of the time, the passenger goes to destination A.. To interpret this statement, we say that the location of the point is X inches from the left-hand-edge of the line, with X L. Since the point is chosen at random on the line, with no further clarification, it is fair to assume that the distribution is uniform, i.e., X has density function f(x = for x L, and f(x = otherwise. L

3 The ratio of the shorter to the longer segment is less than /4 if X L or X 4 L. So 5 5 the ratio of the shorter to the longer segment is less than /4 with probability 5 L L dx + 4. Using proposition 2., we compute E[X n ] = n + = xn+ x n f X (x dx = x= = n + L 4 5 L L (x n ( dx + dx = /5 + /5 = 2/5 (x n ( dx + (x n ( dx = x n dx To use the definition of expectation, we need to find the density of the random variable Y = X n. We first find the cumulative distribution function of Y. We know that P (Y a = for a, and P (Y a = for a. For < a <, we have P (Y a = P (X n a = P (X n a = n a dx = n a. Therefore, the cumulative distribution function of Y is So the density of Y is So the expected value of Y is E[Y ] = = x n + n + n xf Y (x dx = x= = n + 5a. We compute ( X P (X > 5 = P F Y (a = f Y (x = (x( dx + n a if < a < otherwise n a n if < x < otherwise (x n x n dx + (x( dx = n x/n dx > 5 = P (Z > 5/ = P (Z < 5/ Φ( b. We compute ( 4 P (4 < X < = P < X < = P ( < Z < = P (Z < P (Z < = Φ( ( Φ(.82 5c. We compute ( X P (X < 8 = P 5d. We compute ( X 2 P (X < 2 = P < < 8 = P (Z < /3 = P (Z < /3 Φ( = P (Z < 5/3 Φ(

4 4 5e. We compute ( X 2 P (X > = P > = P (Z > = P (Z < = Φ(.587. Let X denote the rainfall in a given year. Then X has a uniform (µ = 4, σ = 4 distribution, so P (X 5 = P ( X = P (Z 2.5 = Φ( , where Z has a standard normal distribution. If the rainfall in each year is independent of all other years, then it follows that the desired probability is (P (X 5 ( We compute ( X 2. = P (X > c = P > c 2 ( = P Z > c 2 ( c 2 = Φ Thus Φ ( c 2 2 =.9. So c So c a. We have n = people, each of which is in favor of a proposed rise in school taxes with probability p =.5. The number of the people who are in favor of the rise in taxes is a Binomial (n =, p =.5 random variable, which is well-approximated by a normal random variable X with mean np = 5 and variance np( p = So the probability that at least 5 are in favor of the proposition is approximately P (X > 49.5 = P ( X > 2b. The desired probability is approximately ( P (59.5 < X < 7.5 = P < X 5 < P (Z > 3.25 = Φ( P (.5 < Z <.5 = P (Z <.5 P (Z <.5 = Φ(.5 ( Φ( c. The desired probability is approximately ( X P (X < 74.5 = P < P (Z <.99 = P (Z < Let X denote the number of times shows during independent rolls of a fair die. Then X is a Binomial random variable with n = and p = /. So X is approximately normal with mean np = / and variance np( p = (/(5/. Thus P (5 X 2 = P (49.5 X 2.5 ( = P 49.5 (/ X (/ (/(5/ (/(5/ P (.4 Z 2.87 = P (Z 2.87 P (Z (/ (/(5/ = P (Z 2.87 P (Z.4 = P (Z 2.87 ( P (Z.4 = Φ(2.87 ( Φ( (.9279 =.9258 Given that shows exactly 2 times, then the remaining 8 rolls are all independent, with possible outcomes, 2, 3, 4, 5, each appearing with probability /5 on each die. Let

5 Y denote the number of times 5 shows during the 8 rolls. Then Y is a Binomial random variable with n = 8 and p = /5. So Y is approximately normal with mean np = 8/5 = and variance np( p = 8(/5(4/5 = 28. Thus ( Y 49.5 P (Y < 5 = P (Y 49.5 = P P (Z.93 = P (Z.93 = P (Z.93 = Φ( = Let X denote the number of acceptable items. Then X is a Binomial random variable with n = 5 and p =.95. So X is approximately normal with mean np = (5(.95 = 42.5 and variance np( p = (5(.95(.5 = Thus ( P (5 X = P (4 X = P (39.5 X = P X P (.2 Z = P (Z.2 = Φ(.2 = With a fair coin, let X denote the number of heads. Then X is a Binomial random variable with n = and p = /2. So X is approximately normal with mean np = 5 and variance np( p = 25. Thus the probability we reach a false conclusion is ( P (525 X = P (524.5 X = P X 5 P (.55 Z = P (Z.55 = Φ( =. With a biased coin, let Y denote the number of heads. Then Y is a Binomial random variable with n = and p =.55. So Y is approximately normal with mean np = 55 and variance np( p = Thus the probability we reach a false conclusion is P (Y < 525 = P (Y < = P ( Y < P (Z.2 = P (Z.2 = P (Z.2 = Φ( = We assume that each person is left-handed, independent of all the other people. Let X denote the number of the 2 people who are lefthanded. Then X is a Binomial random variable with n = 2 and p =.2. So X is approximately normal with mean np = 2(.2 = 24 and variance np( p = 2(.2(.88 = 2.2. Thus the probability that at least 2 of the 2 are lefthanded is ( P (2 X = P (9.5 X = P < X 24 P (.98 < Z = P (Z <.98 = Φ( a. Let X denote the time (in hours needed to repair a machine. Then X is exponentially distributed with λ = /2. Then P (X > 2 = 2 2 e (/2x dx = e b. The conditional probability is P (X > X > 9 = P (X > & X > 9 P (X > 9 = P (X > P (X > 9 = We could also have simply computed the line above by writing P (X > P (X > 9 2 e (/2x dx 9 2 e (/2x dx = e 5 e P (X = P (X 9 = F ( F (9 = ( e (/2 ( e (/29 = e 5 e 9/2 = e /2 5 9/2 = e /2

6 An alternative method is to simply compute that probability that the waiting time would be at least one additional hour, which is P (X > = 2 e (/2x dx = e /2 (or equivalently, P (X > = P (X = F ( = ( e (/2 = e /2. Either way, the answer is e / Let X denote the time in years that the radio continues to functino for Jones. The desired probability is P (X > 8 = 8 e (/8x dx = e (/8x 8 /8 = e.38 8 We could also have written P (X > 8 = P (X 8 = ( e (/88 = e a. We compute P ( X > /2 = P (X > /2 or X < /2 = P (X > /2 + P (X < /2 = /2 + /2 = / b. We first compute the cumulative distribution function of Y = X. For a, we have P (Y a =, since X is never less than a in this case. For a, we have P (Y a =, since X is always less than a in this case. For < a <, we compute x=8 P (Y a = P ( X a = P ( a X a = 2a 2 = a or equivalently, P (Y x = x for < x <. In summary, the cumulative distribution function of Y = X is x F Y (x = x < x < x Differentiating throughout with respect to x yields the density of Y = X is < x < f Y (x = otherwise Intuitively, if X is uniformly distributed on (,, then Y = X is uniformly distribution on [,. 39. Since X is exponentially distributed with λ =, then X has density f X (x = and cumulative distribution function F X (x = e x x otherwise Next, we compute the cumulative distribution function of Y = ln X. We have P (Y a = P (log X a = P (X e a = F X (e a = e ea e x x otherwise or equivalently, P (Y x = e ex. Differentiating throughout with respect to x yields the density of Y = ln X is f Y (x = e x e ex

7 4. Since X is uniformly distributed on (,, then X has density < x < f X (x = otherwise and cumulative distribution function x F X (x = x < x < x Next, we compute the cumulative distribution function of Y = e X. For a, we have P (Y a =, since Y = e X is never less than a in this case (i.e., X is never less than in this case. For a e, we have P (Y a =, since Y = e X is always less than a in this case (i.e., X is always less than in this case. For < a < e (and thus < ln a <, we compute P (Y a = P (e X a = P (X ln a = F X (ln a = ln a or equivalently, P (Y x = ln x for < x < e. In summary, the cumulative distribution function of Y = e X is x F Y (x = ln x < x < e x e Differentiating throughout with respect to x yields the density of Y = e X is f Y (x = x < x < e otherwise 7 THEORETICAL EXERCISES 2. We observe that E[Y ] = xf Y (x dx = xf Y (x dx + xf Y (x dx The second integral can be simplified by observing that x = x dy, so xf Y (x dx = x f Y (x dy dx = y f Y (x dx dy = P (Y > y dy where the second equality in the line above follows by interchanging the order of integration over all x and y with y x. Similarly, x = dy = x dy, so x xf Y (x dx = x f Y (x dy dx = y f Y (x dx dy = P (Y < y dy where the second equality in the line above follows by interchanging the order of integration over all x and y with y x or equivalently x y.

8 8 Altogether, this yields E[Y ] = as desired. P (Y > y dy P (Y < y dy. Let X be uniform on the interval (,. For each a in the range < a <, define ( E a = P (X a. So P (E a = for each a. Also, E a =, so P E a = P ( =. <a< <a< 8. We see that Thus E[X 2 ] = c x 2 f(x dx c cxf(x dx = ce[x] Var(X = E[X 2 ] (E[X] 2 ce[x] (E[X] 2 = E[X](c E[X] = c 2 E[X] c or equivalently, writing α = E[X], we have c Var(X = c 2 α( α ( E[X] c We notice that E[X] = c xf(xdx c E[X] cf(xdx = c, so E[X] c, so, c or equivalently, α. The largest value that α( α can achieve for α is /4, which happens when α = /2 (to see this, just differentiate α( α with respect to α, set the result equal to, and solve for α, which yields α = /2; don t forget to also check the endpoints, namely α = and α =. Since Var(X c 2 α( α and α( α /4, it follows that Var(X c 2 /4, as desired. 2a. If X is uniform on the interval (a, b, then the median is (a + b/2. To see this, write m for the median, and solve /2 = F (m = m m a dx = which immediately yields a b a b a m = (a + b/2. 2b. If X is normal with mean µ and variance σ 2, then the( median is µ. To see ( this, write m X µ for the median, and solve /2 = F (m = P (X m = P m µ = P Z m µ = σ σ σ ( m µ Φ where Z is standard normal, and thus m µ =, so m = µ. σ σ 2c. If X is exponential with parameter λ, then the median is ln 2. To see this, write m λ for the median, and solve /2 = F (m = e m, so e m = /2, so m = ln(/2, so λm = ln 2, and thus m = ln 2. λ 3a. If X is uniform on the interval (a, b, then any value of m between a and b is equally valid to be used as the mode, since the density is constant on the interval (a, b. 3b. If X is normal with mean µ and variance σ 2, then the mode is µ. To see this, write m for the mode, and solve = d dx 2π σ e (x µ2 /(2σ 2, i.e., = 2π σ e (x µ2 /(2σ 2 (2(x µ, 2σ 2 so x = µ is the location of the mode.

9 3c. If X is exponential with parameter λ, then the mode is. To see this, note that d dx λex = λex = e x is the slope of the density for all x >, so the density is always decreasing for x >. So the mode must occur on the boundary of the positive portion of the density, i.e., at x = Consider an exponential random variable X with parameter λ. To show that Y = cx is exponential with parameter λ/c, it suffices to show that Y has cumulative density function e (λ/ca a > F Y (a = otherwise To see this, first we note that X is always nonnegative, so Y = cx is always nonnegative, so F Y (a = for a. For a >, we check F Y (a = P (Y a = P (cx a = P (X a/c = e (a/c = e (λ/ca as desired. Thus, Y = cx has the cumulative distribution function of an exponential random variable with parameter λ/c, so Y = cx must indeed be exponential with parameter λ/c. 8. We prove that, for an exponential random variable X with parameter λ, E[X k ] = k! for k λ k To see this, we use proof by induction on k. For k =, we use integration by parts with u = x and dv = λe x dx, and thus du = dx and v = λex, to see that or, more simply, E[X] = ( λe xλe x x dx = (x x= λe x E[X] = x + e x dx e λx x= We see that the integral evaluates to ex =. We also see that, in the first part of the λ x= expression, plugging in x = yields ; using L Hospital s rule as x yields So we conclude that lim x x = lim eλx x λe λx = E[X] = λ This completes the base case, i.e., the case k =. Now we do the inductive step of the proof. For k 2, we assume that E[X k ] = (k! λ k has already been proved, and we prove that E[X k ] = k!. λ k dx

10 To do this, we use integration by parts with u = x k and dv = λe x dx, and thus du = kx k dx and v = λex, to see that ( λe E[X k ] = x k λe x dx = (x k x ( λe (kx k x dx or, more simply, E[X k ] = xk e λx + k x k λe x dx x= λ We see that the integral is k λ E[Xk ], which is (by the inductive assumption equal to k (k! = k!. We also see that, in the first part of the expression, plugging in x = yields λ λ k λ k ; using L Hospital s rule k times as x yields x k kx k lim = lim x eλx x λe = lim k(k x k 2 k! = = lim λx x λ 2 e λx x λ k e = λx So we conclude that E[X k ] = k! λ k This completes the inductive case, which completes the proof by induction. x=