Notes on Problems for Quiz 5 (10/31/97) from Schaum:
6.36, 6.40, 6.46, 6.47, 6.51, 6.52, 6.55,
6.57, 6.59, 6.60
Problem 6.36:
(1) The probability of not drawing a heart is .75. The probability of not drawing a heart after n tries is .75^n. If we want to make sure that the probability of not drawing a heart is less than x, we must have .75^n < x. Thus, n ln(.75) < ln(x) and n > ln(x)/ln(.75) (the inequality is reversed since ln(.75) is negative). For x = .5, we get n > 2.4. Since n is an integer, n > = 3. Doing the same calculation with x = .25 gives n > 4.8, or n >= 5.
Problem6.40:
(1) The fair die is tossed 1620 times. The probability of a 6 is p = 1/6. The mean is np = 270. The variance is npq = 225. The standard deviation is 15. If we were going to ask for probabilities such as P(200 < X < 250) we would use the normal distribution with this mean and standard deviation. Why not use the Poisson distribution?
Problem 6.46
(1) This is just a matter of using the table of standard normal curve areas on page 111. For each part, you reinterpret the problem in terms of standard deviations from the mean. For (i), the weight of 100 lbs is -2.75 standard deviations below the mean of 155 (100-155 =55 divided by 20). Thus we find the area of the normal curve of page 111 to the left of -2.75. The area from 0 to +2.75 is gotten from the table and is .4970. The area to the right of 2.75 is thus .5 - .4970 = .0030. This is the probability we need for part (i). The number of students with weight below 100 lbs is .003 x 2000 = 6. This is really the "expected number of students" in any real-world application. We will discuss this in class.
Problem 6.47:
(1) Let D denote the random variable, diameter of a bolt. P(D<=.20) = P[(D-.25)/.02 <= (.20-.25)/.02 = -.05/.02 = -2.5] = .5 - .4938 = .0062 from the table on page 111. Similarly, P(D>=.28) = P[(D-.25)/.02 >= 1.5] = .5 - .4332 = .0668. The sum of the two probabilities is .073 which is the expected fraction of bolts that are defective (7.3%).
Problem 6.51: We use the normal distribution with mean np = 120 and variance npq = 100. Thus, for (i), the probability of having between 100 and 125 is the probability of being between -2 and +.5 stamdard deviations from the mean. From the table on page 111, this is .4772 + .1915 = .6687. For (ii), we are interested in the probability of being more than 3 standard deviations from the mean. From page 111, this is .5 - .4987 = .0013.
Problem 6.52
(1) We use the normal approximation to the binomial distribution b(k; 625, .1). The mean is 62.5 and the standard deviation is 7.5. The number 50 is (50 - 62.5)/7.5 = -5/3 standard deviations from the mean.10 is 4/3 standard deviations, so 60 corresponds to -1/3 and 70 to +1 standard deviation from the mean. Now use the table on page 111.
Problem 6.55
(1) Think of the 220 misprints being listed as 1, 2, 3, ... , 220. Fix a page of the book. We assume that any misprint is equally likely to be from this page. Thus, the probability of any given misprint occurring on a particular page is 1/200 = .005. The probability of k misprints on a page is given by the binomial distribution b(k; 220, .005). This is only approximate as we don't really expect the probabilty of 220 misprints on a single page to be the same as the probability of 0 misprints on a single page. This is good approximation for small values of k (k < 10). This is just the situation where we can use the Poisson distribution, here with \lambda = 220/200 = 1.1. The problem asks for values of k=0, 1. We use the table on page 112, or better yet a calculator. p(0; 1.1) = exp(-1.1)=.332 and p(1; 1.1) = .332 x 1.1 = .365 and p(2; 1.1) = (.365 x 1.1)/2 = .201.
Problem 6.57:
(1) Again, this is Poisson approximation to the binomial distribution. In this case, we approximate b(k; 100, .02) with mean 2. In this case exp(-2) = .135.
Problem 6.59:
(1) We use the multinomial distribution of Theorem 6.6 on page 109. Here, n=6, p1=.1, p2=p3=p4=p5=.15, and p6=.30. For (i) we compute 6!/1 x .1 x .15 x .15 x .15 x .15 x .3 = .0109. For (ii) compute 6!/2!2!2! x .15 x .15 x .15 x .15 x .3 x .3 = .004 (answer in book is wrong).
Problem 6.60
(1) The probability of a red marble being drawn is .5, a white is .3, and a blue is .2. We use the multinomial distribution with n = 6. For (i), we compute 6!/3!2!1! x (.5)^3 x (.3)^2 x (.2)^1 = .135. For (ii) the answer is 6!/2!3!1! x (.5)^2 x (.3)^3 x (..2)^1 = .081. For (iii) we get 6!/2!2!2! x (.5)^2 x (.3)^2 x (.2)^2 = .081.