finalexamsolutions1

# finalexamsolutions1 - Math 431 An Introduction to...

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Math 431 An Introduction to Probability Final Exam — Solutions 1. A continuous random variable X has cdf F ( x ) = a for x 0, x 2 for 0 < x < 1, b for x 1. (a) Determine the constants a and b . (b) Find the pdf of X . Be sure to give a formula for f X ( x ) that is valid for all x . (c) Calculate the expected value of X . (d) Calculate the standard deviation of X . Answer: (a) We must have a = lim x →-∞ = 0 and b = lim x + = 1, since F is a cdf. (b) For all x 6 = 0 or 1, F is di±erentiable at x , so f ( x ) = F 0 ( x ) = ( 2 x if 0 < x < 1, 0 otherwise. (One could also use any f that agrees with this deﬁnition for all x 6 = 0 or 1.) (c) E ( X ) = R -∞ x · f ( x ) dx = R 1 0 x · 2 x dx = 2 3 . (d) E ( X 2 ) = R 1 0 x 2 · 2 x dx = 1 2 , Var( X ) = E ( X 2 ) - [ E ( X )] 2 = 1 2 - ( 2 3 ) 2 = 1 18 0 . 0556, and σ ( X ) = q Var( X ) = q 1 18 = 1 3 2 . 2357.

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2. Suppose the number of children born in Ruritania each day is a binomial random variable with mean 1000 and variance 100. Assume that the number of children born on any particular day is independent of the numbers of children born on all other days. What is the probability that on at least one day this year, fewer than 975 children will be born in Ruritania? Answer: We approximate the number of children born each day by a normal random variable. Letting X denote the number of children born on some speciﬁed day, and Z denote a standard normal, we have P ( X 975) = P ( X 974 . 5) = P ( X - 1000 10 974 . 5 - 1000 10 = - 2 . 55) = P ( Z +2 . 55) . 9946. Since each such random variable X (one for each day) is assumed independent of the others, the probability that 975 or more children will be born on every day of this year is . 9946 365 . 1386, and the probability that, on at least one day this year, fewer than 975 children will be born is close to 1 - . 1386 86%. 3. Suppose that the time until the next telemarketer calls my home is distributed as an exponential random variable. If the chance of my getting such a call during the next hour is . 5, what is the chance that I’ll get such a call during the next two hours? Answer: First solution: Letting λ denote the rate of this exponential random variable X , we have . 5 = F X (1) = 1 - e - λ , so λ = ln2 and F X (2) = 1 - e - 2 λ = 1 - ( e - λ ) 2 = 1 - ( . 5) 2 = . 75. Second solution: We have P ( X 2) = P ( X 1) + P (1 < X 2). The ﬁrst term is . 5, and the second can be written as P ( X > 1 and X 2) = P ( X > 1) P ( X 2 | X > 1).
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## This note was uploaded on 04/24/2011 for the course ECE 6303 taught by Professor Voltz during the Spring '11 term at NYU Poly.

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finalexamsolutions1 - Math 431 An Introduction to...

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