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University of Washington_MATH 394-HW5.solution

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HOMEWORK 5: SOLUTIONS Abstract. The problems are taken from the textbook. The numbers are referred to in the beginning. You do not need to use continuity correction for the exams. So, you can ignore those comments in t ...

HOMEWORK 5: SOLUTIONS Abstract. The problems are taken from the textbook. The numbers are referred to in the beginning. You do not need to use continuity correction for the exams. So, you can ignore those comments in the following solutions. Problem 1. We choose 500 numbers uniformly at random from the interval [1:5; 4:8]. (a) Approximate the probability of the event that less than 65 of the numbers start with the digit one. (b) Approximate the probability of the event that more than 160 of the numbers start with the digit 3. Solution. You do not need to use continuity correction for the exams. Date: Nov 7, 2018. 12 = ( ([0, ]) = 0, ([ , ]) = 13) P(N([0, 4]) = 13) = P(N([0, 1]) = 0) · P(N([1, 4]) = 13) P(N([0, 4]) = 13) = e!4 · e!121213/13! e!161613/13! = ✓3 4◆13 ⇡ 0.02376. 4.16. (a) Denote by S the number of random numbers starting with the digit 1. Note that a number in the interval [1.5, 4.8] starts with 1 if and only if it is in the interval [1.5, 2). The probability that a uniformly chosen number from the interval [1.5, 4.8] is in [1.5, 2) is equal to p = 4.80!.51.5 = 33 5 . Assuming that the 500 numbers are chosen independently, the distribution of S is binomial with parameters n = 500 and p. To estimate P(S < 65) we use normal approximation. Note that E[S] = np = 500 · 33 5 ⇡ 75.7576 and Var(S) = np(1 ! p) ⇡ 64.2792. Hence P(S < 65) = P ✓Sp!6475 .2792 .7576 < 65p!6475 .2792 .7576◆ ⇡ P ✓Sp!6475 .2792 .7576 < !1.34◆ ⇡ !(!1.34) = 1 ! !(1.34) ⇡ 1 ! 0.9099 = 0.0901. Note that P(S < 65) = P(S  64). Using 64 instead of 65 in the calculation above gives 1 ! !(1.47) ⇡ 0.0708. If we use the continuity correction then we Solutions to Chapter 4 97 need to use 64.5 instead of 65 which gives 1 ! !(1.4) ⇡ 0.0808. The actual probability (evaluated numerically) is 0.0778. (b) We proceed similarly as in part (a). The probability that a given uniformly chosen number from [1.5, 4.8] starts with 3 is q = 31.3 = 10 33. If we denote the number of such numbers among the 500 random numbers by T then T ⇠ Bin(n, q) with n = 500. Then

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