Question

General Mills (GM) manufacturers and sells a wide variety of breakfast cereals. GM’s product development lab recently created a new cereal that consist of rice flakes and banana-flavored marshmallows. The company’s marketing research department has tested the new cereal extensively and has found that consumers are enthusiastic about the cereal when 16 ounce boxes contain at least 1.6 ounces and no more than 2.4 ounces of the banana-flavored marshmallows. As GM prepares to begin producing and selling 16-ounce boxes of the new cereal, which it has named Cheerios, management is concerned about the amount of banana-flavored marshmallows. It wants to be careful not to include less than 1.6 ounces or more than 2.4 ounces of bananaflavored marshmallows in each 16-ounce box of Cheerios. Tina Finkel, VP of Production for GM, has suggested that the company measure the weight of banana-flavored marshmallows in a random sample of 25 boxes of Cheerios on a weekly basis. Each week GM can count the number of boxes out of the 25 boxes in the sample that contain less than 1.6 ounces or more than 2.4 ounces of banana-flavored marshmallows; if the number of boxes that fail to meet the standard weight of banana-flavored marshmallows is too high, production will be shut down and inspected. Ms. Finkel and her staff have designed the production process so that only 8% of all 16-ounce boxes of Cheerios fail to meet the standard weight of banana-flavored marshmallows. After much debate, GM management has decided to shut down production of Cheerios if at least five boxes in a weekly sample fail to meet the standard weight of banana-flavored marshmallows.

Calculate the probability that a weekly sample will result in a shutdown of production if the production process is working properly. Comment on GM management’s policy for deciding when to shut down production of Cheerios. 2. GM management wants to shut down production of Cheerios no more that 1% of the time when the production process is working properly. Suggest the appropriate number of boxes in the weekly sample that must fail to meet the standard weight of banana-flavored marshmallows in order for production to be shut down if this goal is to be achieved. 3. Ms. Finkel has suggested that if given sufficient resources, she could redesign the production process to reduce the percentage of 16-ounce boxes of Cheerios that fail to meet the standard weight of banana-flavored marshmallows when the process is working properly. To what level must Ms. Finkel reduce the percentage of 15-ounce boxes of Cheerios that fail to meet the standard weight of banana-flavored marshmallows when the process is working properly in order for her to reduce the probability at least five of the sampled boxes fail to meet the standard to .01 or less?

Answer 1

This is a case of binomial experiment consisting of n trials and results in x successes. If the probability of success on an individual trial is p, then the binomial probability is: P(X = x) = (*) p*(1- p)** From the information, n=25, p=0.08 The Probability that a weekly sample will result in a shutdown of production if the production process is working properly is the probability that at least 5 boxes out of 25 samples fail to meet the standard weight P(X 25)=1-P(X<5) 11 ---Ź[23* 10.08*(1-0.05)*7} 1620.08°(1–0.08)350 +(25 0.08' (1-0.08)25- =1- +(39)0.08°(1–0,05).*2 = (35 0.08°(1 – 0.08)*5 +(2510.08*(1-0.08)234 =1-{0.1244 +0.2704+0.2821+0.1881+0.0899} = 1-0.9549 = 0.0451 So, probability that a weekly sample will result in a shutdown of production if the production process is working properly is 4.51%.

The probability that a weekly sample will result in a shutdown of production if the production process is working properly is 0.01 Now find B (number of boxes) such that, P(X2x)= 25 p*(1- p)** 30.01 If B=5, P(X 25)=1-P(X<5) -1-{$(23)0.08*(1-0.08)*} 250.08°(1–0.08)*50 +(25 0.08' (1-0.08)*** =- +(29).08°(1–0.05)*2 (35 0.8°(1 – 0.08)** +(23)0.08*(1 – 0.08)*** = 1-0.1244 +0.2704+0.2821+ 0.1881+0.0899) = 1-0.9549 = 0.0451

If B=6, P(X26)=1-P(X<6) . =1-Ź(33)0.08* (1 – 0.08)* 1635 10.08°(1 – 0.08)** +(35 0.08*(1–0,03/43 =1+2+(29)0.08° (1 +0.08)352 +(39).08*(1 0.08)*** +(23)0.08* (1–0.08)*** +(33)0.08*(1 0.08)*** 15 =1-0.1244+0.2704 +0.2821+0.1881+0.0899+0.0329 =1-0.9877 = 0.0123 If B= 7, P(X27)=1-P(X<7) 16 (25 x=0 =1-30770.08*(1 – 0.08)*** |(2510.08°(1–0.08) 250 +(25.0.08(1 – 0.08)234 +(23)0.08(1–0.08)2+(35)0.08*(1-0.08)*** +6290.08*(1-0.08)3-4 +(25 0.08*(1-0.08) 3- (250.08*(1-0.08)256 + + + 16) 10.1244+0.2704 +0.2821+0.1881 =1- | +0.0899 +0.0329 +0.0095 =1-0.9972 = 0.0028 <0.01

So at least 7 boxes in the weekly sample must fail to meet the standard weight for production to be shut down so that the probability that a weekly sample will result in a shutdown of production if the production process is working properly is less than 1%. 3. Find the probability that a sample fail to meet the standard weight of banana-flavoured marshmallows if the production process is working such that probability that a weekly sample will result in a shutdown of production if the production process is working properly is less than 1% or Probability that at least 5 boxes out of 25 samples fail to meet the standard weight is less than 1% that is P(X 25) s0.01. From previous part for p=0.8 it is observed that P(X 25)=0.0452 Using binomial simulation, we determine that P(x 5) s0.01 for p=0.0542 So, the production process should be redesigned to reduce the percentage of 16-ounce boxes of Go Bananas! that fail to meet the standard weight of banana-flavoured marshmallows if the production process is working properly to 5.42%