Group Project2-Answers-6November2020 - Tagged

Group Project 2 Solution 50 Points Due by 11:59 PM EST on Sunday October 25, 2020 1. (10 points) The tolerance limits for a particular quality characteristic (e.g., length, weight, or strength) of a particular product are the minimum and/or maximum values at which the product will operate properly. Tolerance limits are set by the engineering design function of the manufacturing operation. The tensile strength of a particular metal part can be characterized as being normally distributed with a mean of 25 pounds and a standard deviation of 2 pounds. The upper and lower tolerance limits for the part are 30 pounds and 21 pounds respectively. A part that falls within the tolerance limits results in a profit of $10. A part that falls below the lower tolerance limit costs the company $2; a part that falls above the upper tolerance limit costs the company $1. Find the company’s expected profit per metal part produced. X: tensile strength of a metal part (in pounds) X ~ N(25, (2) 2 ) The lower and upper tolerance limits are 21 pounds and 30 pounds respectively. Y: the profit per metal part produced. Then, the probability distribution of Y is Y ($) P(Y) 10 P(21<x<30) -2 P(x<21) -1 P(x>30) P(Y=10) = P(21<x<30) = P ¿ < x − 25 2 < 30 − 25 2 ) = P ¿ < 2.5 ) ¿ P ¿ + P(z<-2) = NORM.S.DIST(2.5,TRUE)- NORM.S.DIST(-2,TRUE) = 0.97104 ≈ 0.9710 . P(Y=-2) = P(X<21) = P ¿ < 21 − 25 2 ) = P ¿ < − 2 ) ¿ P ¿ =1-NORM.S.DIST(2,TRUE) = 0.02275 ≈ 0.0228 P(Y=-1) = P(x>30) = P ¿ > 30 − 25 2 ) = P ¿ > 2.5 ) ¿ 1-NORM.S.DIST(2.5,TRUE) = 0.00621 ≈ 0.0062 So, E(Y) = (10)(0.9710) + (-2)(0.0228) + (-1)(0.0062) = $9.6582.

2. (10 points) The Globe Fishery packs shrimp that weigh more than 1.91 ounces each in packages marked" large" and shrimp that weigh less than 0.47 ounces each into packages marked "small"; the remainder are packed in "medium" size packages. If a day's catch showed that 19.77 percent of the shrimp were large and 6.06 percent were small, determine the mean and the standard deviation for the shrimp weights. Assume that the shrimps' weights are normally distributed. 0.1977 shrimps were large -- more than 1.91 ounces in weight z =NORM.S.INV(0.1977) = -0.84987. Since we are looking for positive z, we will consider |-0.84987|= 0.84987 ≈ 0.85 . We know that, 0.85 = 1.91 − μ σ Or, 1.91 - μ = 0.85 σ Or, μ = 1.91 − 0.85 σ (1) 0.0606 shrimps were small -- less than 0.47 ounces in weight z =NORM.S.INV(0.0606) = -1.5498 ≈ − 1.55 . We know that, -1.55 = 0.47 − μ σ Or, 0.47 - μ =− 1.55 σ Or, μ = 0.47 + 1.55 σ (2) (1) = (2) implies 1.91 − 0.85 σ = 0.47 + 1.55 σ Or, 1.91 – 0.47 = 1.55 σ + ¿ 0.85 σ Or, 2.4 σ = 1.44 Or, σ = 1.44 2.4 = 0.6 Substituting σ ∈ ( 2 ) ,we get μ = 0.47 + 1.55 ( 0.6 ) = 0.47 + 0.93 = 1.40 Note: We would get the same μ if we would have plugged σ ∈ ( 1 ) . Thus, our μ and σ are 1.4 ounces and 0.6 ounce respectively.

In a survey of MBA students, the following data were obtained on “students’ first reason for application to the school in which they matriculated”. Reason for Application Enrollment Status School Quality School Cost or Convenienc e Other Full Time 421 393 76 Part Time 400 593 46 4. (10 points) Suppose that one of the students is selected at random. Using this data set, answer the following questions. a. Develop a joint probability table for this data. Reason for Application Enrollment Status School Quality School Cost or Convenience Other Total Full Time 421 393 76 890 Part Time 400 593 46 1039 Total 821 986 122 1929 Quality Cost Other Total Full Time 0.2182 0.2037 0.0394 0.4613 Part Time 0.2074 0.3074 0.0238 0.5386 Total 0.4256 0.5111 0.0632 1.0000 b. Use the marginal probabilities of school quality, school cost or convenience, and other to comment on the most important reason for choosing a school. Marginal Probability of Quality: P(Q) = 0.4256 → 42.56% of students chose school based on quality. Marginal Probability of Cost: P(C) = 0.5111 → 51.11% of students chose school based on cost or convenience. Marginal Probability of Other: P(O) = 0.0632 → 6.32% of students chose school based on other considerations. Comparing the three factors, we see that cost was the most important reason for choosing a school.

c. If a student goes full time, what is the probability that school quality is the first reason for choosing a school? P(Q|FT) = P ( Q∩ FT ) P ( FT ) = 0.2182 0.4613 = 0.4730 d. If a student goes part time, what is the probability that School Cost or Convenience is the first reason for choosing a school? P(C|PT) = P ( C∩ PT ) P ( PT ) = 0.3074 0.5386 = 0.5707 e. Let A denote the event that a student is full time and let B denote the event that the student lists school quality as the first reason for applying. Are events A and B independent? Justify your answer. 2 events A and B are independent if P(A|B) = P(A) and/or P(B|A) = P(B). We can work with one of the two conditions. In our example, P(Q|FT) = 0.4730 and P(Q) = 0.4256. Since the two probabilities are not equal, the two variables are dependent. This means that quality is affected by whether the student is full time or part time. 5. (6 points) A box consists of 21 toffees of which 12 are green and 9 are blue. You picked 2 toffees at random. Find the probability that both toffees are blue. =HYPGEOM.DIST(2,2,9,21,FALSE) = 0.171429 ≈ 0.1714 N = 21 total number of toffees g = 12 green toffees in N = 21 toffees; 21-12 = 9 toffees are blue, that is b = 9. n = 2 toffees are drawn at random. x = number of blue toffees in the n = 2 toffees drawn without replacement. Another way of solving the problem: P(2 Blue Toffees) = P(B ∩ B ¿ = 9 21 ∗ 8 20 = 0.17142857 ≈ 0.171429 ≈ 0.1714 (since the drawings are without replacement).