Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
ENGR 323 Spring 1998 HW #14 Heidi M. Gehlhaar May 1, 1998 1/2 Problem 5-97 5-97) Problem Statement: The weight of a small candy is normally distributed with a mean of 0.1 ounce and a standard deviation of 0.01 ounce. Suppose that 16 candies are placed in a package and that the weights are independent. a.) What are the mean and variance of package net weight? b.) What is the probability that the net weight of a package is less than 1.6 ounces? c.) If 17 candies are placed in each package, what is the probability that the net weight of a package is less than 1.6 ounces? Problem Solution: We will begin with variable definition: Let Wi = Weight of the ith small candy in ounces Wi ~ N(µW = 0.1, σW = 0.01) Let X = Weight of a package in ounces containing 16 candies Because X is just 16 of W, X is just a linear combination of W and is normally distributed and has the same type of random variables with different parameters: X = W1 + W2 + W3 + ... + W16 Wi are iid X ~ N(µX = ?, σX = ?) a.) Here we will find the parameters of the distribution of X, including the mean and variance. To do this, we will use the general rule we learned in class about linear combinations of random variables (page 273): For A = c1B1 + c2B2 +...+ cnBn, E(A) = c1E(B1) + c2E(B2) +...+ cnE(Bn) V(A) = c12(B1) + c22(B2) + ...+ cn2(Bn) For our problem, E(X) = E(W1) + E(W2) + ... + E(W16) = 16(0.1) E(X) = 1.6 ounces V(X) = V(W1) + V(W2) + ... + V(W16) = 16(0.012) V(X) = 0.0016 ounce where E(W) = µW = 0.1 where V(W) = σW2 = 0.012 ENGR 323 Spring 1998 HW #14 Heidi M. Gehlhaar May 1, 1998 2/2 Problem 5-97 Continued It is very important to note that we use the variance of the random variables, not the standard deviation! Many mistakes can be made by accidentally using the standard deviation instead of the variance in a linear transformation. Beth pointed out that it is just as important to remember as a2+b2=c2 and NOT a+b=c for a right triangle! The two concepts are very similar and equally messed up by unknowing students. The expected value can be changed linearly but the standard deviation cannot. When performing problems such as this one, be sure to note if your standard deviation changed by a reasonable amount. When doing a linear combination of random variables, sigma will increase, but not as much as it would if increased linearly with the expected value. An extreme change in expected value will definitely give you a change in standard deviation but not as dramatically as might be expected. b.) The probability that a package weighs less than 1.6 ounces can be written as P(X<1.6). Since we have already shown that X is distributed normally, we can standardize and use our Z table to solve this portion of the problem. 1.6 − 1.6 P(X<1.6) = Z < 0.04 = P(Z<0) P(X< 1.6) = 0.5 50% makes sense seeing as the problem was really asking for the probability that a package weighs less than the mean. We would expect to find packages that weigh less than the mean half of the time and packages that weight more than the mean the other half of the time. (Remember that, with continuous random variables, P(X=x) = 0.) c.) This portion of the problem is a combination of parts a and b but with packages containing 17 candies instead of 16 candies. We need a new random variable: Let Y = Weight of a package in ounces containing 17 candies Y = W1 + W2 + W3 + ... + W17 Wi are iid As before, E(Y) = E(W1) + E(W2) + ... + E(W17) where E(W) = µW = 0.1 = 17(0.1) E(Y) = 1.7 ounces And V(Y) = V(W1) + V(W2) + ... + V(W16) where V(W) = σW2 = 0.012 = 17(0.012) V(Y) = 0.0017 ounce As explained earlier, the distribution of Y will be normal: Y ~ N(µY = 1.7, σY = 0.0412) Now we find the desired probability by standardizing: 1.6 − 1.7 P(Y<1.6) = Z < 0.0017 = P(Z < -2.43) Then use the Z table: P(Y<1.6) = 0.008