CMPE 107 - Homework 6 due Feb. 27 1. The random variable x is normal, having pdf equal to gµ ,! ( x ) with µ=2. a. What is P(x ≤ 2)? b. Knowing that P(x > 4)=0.16, can you compute P(x ≤ 0)? c. Knowing that gµ ,! (4 ) = 0.12 , can you compute gµ ,! (0 ) ? d. What’s P(2< x ≤4)? • • • • P(x≤2) = G2,σ(2) = G((2-2)/σ) = G(0) = 0.5 P(x > 4) = 1 – P(x ≤ 4) = 1 - G2,σ(4) = 1 - G((4-2)/σ) = 1 - G(2/σ) P(x ≤ 0) = G2,σ(0) = G((0-2)/σ) = G(-2/σ) Due to the symmetry of G(x), we have that G(-2/σ) = 1 - G(2/σ) = P(x > 4) = 0.16 g2,σ(4) = g((4-2)/σ) = g(2/σ) g2,σ(0) = g((0-2)/σ) = g(-2/σ). Due to the symmetry of g(x), we have g(-2/σ) = g(2/σ)= g2,σ(4) = 0.12 P(2< x ≤4) = G2,σ(4) - G2,σ(2) = (see above) = (1 - P(x ≤ 4)) – 0.5 = (1-0.16)-0.5=0.34 2. The CDF of a random variable x is 0 x < !5 $ & ( x + 5) / 8 !5 " x < !3 Fx ( x ) = % 1/ 4 !3 " x < 3 &1/ 4 + 3(x ! 3) / 8 3 " x < 5 1 x #5 ' a. What is P(x ≤ 4)? b. What is P(-2< x ≤2)? c. What is P(x > 0)? d. What is the value of a so that P(x ≤ a) = 0.5? • • • • Fx(4) = 5/8 Fx(2)-Fx(-2) = 1/4 – 1/4 = 0 1 – Fx(0) = 3/4 x = 11/3 (note that this is by definition the median) 3. The cumulative distribution function of the random variable x is 0 x < !1 $& Fx ( x ) = %(x + 1)/ 2 !1 " x < 1 &' 1 x#1 Find the pdf fx(x) of x. x < !1 $& 0 f x ( x ) = %0.5 !1 " x < 1 &' 0 x #1 therefore it is uniform. 4. Let x be a random variable with cdf 0 x < !1 # % x / 3 + 1/ 3 !1 " x < 0 Fx ( x ) = $ x / 3 + 2/ 3 0 " x < 1 % 1 1" x & Sketch the cdf and find a. P(x < -1) and P(x ≤ -1) b. P(x < 0) and P(x ≤ 0) c. P(0< x ≤1) and P(0 ≤ x ≤1) • • • P(x ≤ -1) = Fx(-1) = 0; P(x < -1) = P(x ≤ -1) because Fx(x) is continuous in –1. P(x ≤ 0) = Fx(0) = 2/3; P(x < 0) = P(x ≤ 0) – P(x=0). Notice that P(x=0)=1/3 because Fx(x) is discontinuous in that point. Therefore, P(x < 0) = 1/3 P(0 < x ≤1) = Fx(1) – Fx(0) = 1/3; P(0 ≤ x ≤1) = P(0 < x ≤1) + P(x=0) = 2/3 5. The peak temperature T, in degrees Fahrenheit, of a July day in Antarctica is a Gaussian random variable with a variance of 225. With probability 0.5, the temperature T exceeds 10 degrees. a. What is P(T>32), the probability the temperature is above freezing? b. What is P(T<0)? c. What is P(T>60)? The variance is σ2225. Therefore, σ=15. It is easy to show that the median of a Gaussian random variable is µ (this is because Gµ,σ(µ) = G(0) = 0.5) Therefore, µ=10. (Remember that the median of a RV is the value for which P(x≤0) = P(x>0) = 0.5) • 1-G10,15(32) = 1-G(22/15) • G10,15(0) = G(-10/15) • 1-G10,15(60) = 1-G(50/15) 6. The random variable x has probability density function "cx 0 ! x ! 2 fx ( x) = # $ 0 otherwise Use this pdf to find: a. b. c. d. 2 ! x2 $ cx dx = 1 = c '0 #" 2 &% = 2c . Therefore c = 0.5. 0 2 • 1 x " x2 % !0 2 dx = $# 4 '& 0 = 0.25 # x 2 &0.5 1 0.5 x " 0 2 dx = %$ 4 (' = 16 0 x2/4 for 0≤x≤2; 0 for x<0; 1 for x ≥2. 1 • • • ! The constant c P(0 ≤ x ≤1) P(-0.5 ≤ x ≤0.5) The CDF Fx(x) 7. The random variable x has pdf "ax 2 + x 0 ! x ! 1 fx ( x) = # otherwise $ 0 What condition on the constant a is necessary and sufficient to guarantee that fx(x) is a valid pdf? 1 " x3 x2 % a 1 !a ax + x = $#a 3 + 2 '&0 = 3 + 2 = 1 therefore a=3/2. Note that fx(x) is always ≥ 0. 1 2 8. The peak temperature T, as measured in degrees Fahrenheit, on a July day in Los Angeles, is a Gaussian random variable with µ=85 and σ=10. a. What is P(T>100),? b. What is P(T<60)? c. What is P(70<T≤100)? • • • 1-G85,10(100) = 1-G(1.5) = 0.07 G85,10(60) = G(-2.5) = 1-G(2.5) = 0.01 G85,10(100) - G85,10(70) = G(1.5) – G(-1.5) = 2 G(1.5) – 1 = 0.87