HW3

1. Suppose a continuous random variable X has the following probability density function (a) Find the value of k that makes fX(x) a valid probability density function. (Recall a property that a PDF must have) f X (x)  0 for all x, we have x(1-x)  0 for all 0  x  1 => k  0 (1) ∫ − ∞ ∞ fX ( x ) dx = 1 => ∫ − ∞ 0 0 dx + ∫ 0 1 kx ( 1 − x ) dx + ∫ 0 ❑ 0 dx = 1 => ∫ 0 1 kx ( 1 − x ) dx = 1 => ∫ 0 1 kxdx - ∫ 0 1 k x 2 dx = 1 => kx 2 /2| 1 0 – kx 3 /3| 1 0 = 1 => k ( 1/2 - 1/3 ) = 1 => 1/6 k = 1 => k = 6 (2) (1) and (2) => k = 6 (b) Give the CDF, F X (x). F X (x) = P (X  x) For x < 0: F X (x) = 0 For x > 1: F X (x) = 1 For 0  x  1: F X (x) = ∫ 0 x 6 x ( 1 − x ) dx = 3x 2 | x 0 – 2x 3 | x 0 = 3x 2 – 2x 3 F X (x) = { 0 ,x < 0 3 x 2 – 2 x 3 , 0 x 1 1 ,x > 1

(c) Find P(0.5 ≤ X ≤ 1) using f X (x). P(0.5 ≤ X ≤ 1) = ∫ 0.5 1 6 x ( 1 − x ) dx = 0.5 (d) Find P(0 ≤ X ≤ 0.75) using F X (x). P(0 ≤ X ≤ 0.75) = F X (0.75) – F X (0) = 3 ( 0.75 ) 2 – 2 ( 0.75 ) 3 – 0 = 0.84375 (e) Find E(X). E(X) = ∫ 0 1 6 x 2 ( 1 − x ) dx = ∫ 0 1 6 x 2 dx - ∫ 0 1 6 x 3 dx = 2x 3 1 0 - 3/2 x 4 1 0 = 0.5 (f) Find Var(X). E(X 2 ) = ∫ 0 1 6 x 3 ( 1 − x ) dx = ∫ 0 1 6 x 3 dx - ∫ 0 1 6 x 4 dx = 3/2x 4 1 0 - 6/5 x 5 1 0 = 3/10 Var(X) = E(X 2 ) – E(X) 2 = 3/10 – (1/2) 2 = 0.05 2. Suppose a continuous random variable X has the following probability density function: (a) Write down the cumulative distribution function (CDF) in functional form. (Make sure to cover all cases). F X (t) = P (X  t) For t < -1: F X (t) = 0 For t > 1: F X (t) = 1 For -1  t  1: F X (x) = ∫ − 1 t 3 2 x 2 dx = 1 2 x 3 | t − 1 = 1 2 (t 3 +1) F X (t) = { 0 ,t ← 1 1 2 ( t 3 + 1 ) , − 1 t 1 1 ,t > 1

(b) Find P(0.5 ≤ X ≤ 1.2). P(0.5 ≤ X ≤ 1.2) = F X (1.2) – F X (0.5) = 1.2 – ½ - (0.5) 2 /2 = 0.575 (c) Find E(X) E(X) = ∫ 0 1 x 2 dx + ∫ 1 1.5 xdx = 0.3333 + 0.625 = 0.9583 4. You arrive at a bus stop at 10:00, knowing that the bus will arrive at some time uniformly distributed between 10:00 and 10:30. Let X = time you wait for the bus to arrive (in minutes). Thus, X ∼ Unif(0; 30) (a) How many minutes do you expect to wait? E(X) = ( 0+30) /2 = 15 minutes (b) What is the probability that you have to wait longer than 10 minutes? F X (t) = { 0 ,t < 0 t 30 , 0 t 30 1 ,t > 30 P(X>10) = 1 – P(X  10) = 1 - F X (10) = 1 – 10/30 = 2/3 = 0.6667 (c) If at 10:15 the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes? P(15 < X  25) = F X (25) – F X (15) = 25/30 – 15/30 = 1/3 = 0.33333 (d) Suppose you go grab a coffee right when you get to the stop at 10:00. What time do you need to be back at the stop, such that there is only a 10% chance you will miss the bus? P(0<X<t) = 0.1 => F X (t) - F X (0) = 0.1 => F X (t) = 0.1 => t/30 = 0.1 => t = 3 minutes You need to be back at the stop at 10:03 5. A web page is accessed at an average of 20 times an hour. Assume that waiting time until the next hit has an exponential distribution.

(a) Determine the rate parameter λ of the distribution of the time until the first hit?  = 20 , X = time between access , X ~ Exp(20) F X (t) = { 0 for t < 0 1 − e − 20 t for t 0 (b) What is the expected time between hits? 1/  = 1/20 = 0.05 h (c) What is the probability that the next hit is within 20 minutes? P(X  1/3) = F X (1/3) = 1 − e − 20 3 = 0.9987 (d) What is the distribution of the time until the second hit? (Give the name of the distribution and the value(s) of parameter(s).) X ~ Gamma(2,20) (e) Describe the distribution of the total waiting time for 5 hits? (Give the name of the distribution and the value(s) of parameter(s).) X ~ Gamma(5,20) (f) What is the expected total waiting time for 5 hits on the web page? E(X) =  /  = 5/20 = 0.25 (g) What is the probability that there will be less than 5 hits in the first hour? (Hint: Consider Poisson distribution instead.) X ~ Gamma(5,20) and T ~ Pois(20) P(X>1) = P(T<5) = P(T  4) = F T (4) = 0.00017 6. In lecture, we talked about the memoryless property of the exponential distribution. Let Y ∼ Exp(λ). For t > s, we stated that P(Y ≤ s + t| Y ≥ s) = P(Y ≤ t) Prove that this is true P(Y ≤ t) = F Y (t) = 1 – e -  t P(Y ≤ s + t| Y ≥ s) = P ( Y ≤s + t ) P ( Y s ) P ( Y s ) = P ( sY s + t ) 1 − F Y ( s ) = F Y ( s + t ) − F Y ( s ) 1 − F Y ( s ) = 1 − e − s − t − 1 + e − s 1 − 1 + e − s = e − s ∗( 1 − e − t ) e − s = 1 – e -  t Therefore, P(Y ≤ s + t| Y ≥ s) = P(Y ≤ t)

P(T<5) = P(X  25) = 1 – P(X < 25) = 1 – P(X  24 ) = 0.8428 9. For a Normal random variable X with E(X) = −3 and Var(X) = 4, Compute (a) P(X ≤ 2.39) P(X  2.39) = P ( X + 3 2 2.39 + 3 2 ¿ = P(Z  2.695) = 0.9964 (b) P((X+3)/2 ≤ −0.99) P((X+3)/2 ≤ −0.99) = P(Z ≤ −0.99) = 0.1611 (c) P(|X| ≥ 2.39) P(|X| ≥ 2.39) = P (X  -2.39) + P(X  2.39) = P ( X + 3 2 0.305 ¿ + P ( X + 3 2 2.695 ¿ = P(Z  0.305)+P(Z  2.695) = 0.6198 + 0.0035 = 0.6233 (d) P(|X + 3| ≥ 2.39) P(|X + 3| ≥ 2.39) = P(X + 3 ≥ 2.39) + P(X + 3  -2.39) = P ( X + 3 2 − 1.195 ) + P ( X + 3 2 1.195 ¿ =P(Z − 1.195 ¿ + P(Z 1.195 ¿ = 0.116 + 0.116 = 0.232 (e) P(X < 5) = P ( X + 3 2 < 4 ¿ = P(Z < 4) = 1 (f) P(|X| < 5) = P(-5<X<5) = P(-1< X + 3 2 <4) = P(-1<Z<4) = 0.8413 (g) the value x such that P(X > x) = 0.33 P( X + 3 2 > x + 3 2 ) = 0.33 => P(Z > x + 3 2 ) = 0.33 => x + 3 2 = 0.44 => x = -2.12 10. The price of a particular make of a 64GB iPad Mini among dealers nationwide is assumed to have a Normal distribution with mean µ = $500 and variance σ 2 = 225.  = 15, X = Price of the 64GB iPad Mini (a) What is the probability that an iPad Mini of the same specs, chosen randomly from a dealer, will cost less than $490? P(X<490) = P( X − 500 15 < -2/3) = P(Z<-0.67) = 0.2514

(b) What is the probability that an iPad Mini of the same specs, chosen randomly from a dealer, will cost more than $530? P(X>530) = P( X − 500 15 > 2) = P(Z >2) = 1 – P(Z  2) = 1 – 0.9772 = 0.0028 (c) What is the probability that an iPad Mini of the same specs, chosen randomly from a dealer, will cost between $490 and $530? P(490<X<530) = P(-0.6667 < Z < 2) = P(Z<2) – P(Z<-0.6667) = 0.9772 – 0.2514 =0.7258 (d) The manufacturer doesn’t want dealers to markup these iPad Minis above the 90th percentile of the price distribution. Approximately, what is the 90th percentile of the distribution of the price of this iPad Mini? P(Z  P) = 0.9 => P =1.28 => X − 500 15 = 1.28 => X = 519.2 (e) What is the probability that the average price of 25 iPad Minis of the same specs, chosen randomly from a dealer, will be greater than $510?  average =  = 500 X average = 1/25  X  2 Xaverage = 1/25 * 225 = 9 P(X average > 510) = P(Z > 510 − 500 3 ) = P(Z>3.33) = 1 – P(Z  3.33) = 1 – 0.9996 = 0.0004 11. Suppose the installation time in hours for a software on a laptop has probability density function f(x) = 4/3 (1 − x 3 ), 0 ≤ x ≤ 1. (a) Find the probability that the software takes between 0.3 and 0.5 hours to be installed on your laptop. P(0.3  X  0.5) = ∫ 0.3 0.5 4 3 ( 1 − x 3 ) dx = 0.2485 (b) Let X1; : : : ; X30 be the installation times of the software on 30 different laptops. Assume the installation times are independent. Find the probability that the average installation time is between 0.3 and 0.5 hours. Cite the theorem you use. Central Limit Theorem n = 30  = E(X) = ∫ 0 1 4 3 x ( 1 − x 3 ) dx = 2/5

P(S n < 1200) = P(Z < 1200 − 82 ∗ 15 4 ∗ √ 82 ) = P(Z < -0.8282) = 0.2038 13. Extra Credit (2 points) Let X be a continuous random variable with CDF FX(t). Let U ∼ unif(0; 1). Define a random variable Y as Y = F −1 X (U). Show that the CDF of Y is the same as the CDF of X, i.e. show FY (t) = P(Y ≤ t) = …. = FX(t). Fill in the dots to show this. F Y (t) = P (Y  t) = P (F −1 X (U)  t ) = P (U  f X (t)) = ∫ 0 fX ( t ) 1 1 − 0 dx = ∫ 0 fX ( t ) dx = x fX ( t ) 0 = F X (t)