0000 1s 00000 0.0000 2 0001 0000 0011 0.000% 0010 Problem 1: Filling Tide Deterg

Question

0000 1s 00000 0.0000 2 0001 0000 0011 0.000% 0010 Problem 1: Filling Tide Detergent Bottles Proctor and Gamble (P&G;) manufactures liquid Tide detergent (among many other products). Liquid Tide is sold in plastic bottles. One of the final steps in the manufacturing process is to fill the bottles of Tide. One machine used to fill the bottles is set to put an average of 100 ounces of Tide in each bottle. However, this machine cannot be guaranteed to put exactly 100 ounces of Tide in each bottle. Rather, the fill amount is known to follow a normal distribution with mean of 100 ounces and standard deviation of 0.2 ounces. Thus, some bottles will contain slightly more than 100 ounces and some slightly less, even though these bottles will be labeled as 100 ounce" bottles. a. What is the probability that less than 99.6 ounces will be put into a "100 ounce" bottle of liquid Tide? b. Calculate the probability that a single bottle of Tide will contain between 99.9 and 100.1 ounces. c. What is the 90th percentile of the fill amounts? d. Suppose P&G; can adjust the mean fill amount on the machine that fills the Tide bottles. At what value should the mean-fill be set to ensure that at most 5% of the Tide bottles will contain less than 99.8 ounces?

Explanation / Answer

mean = 100 , s = 0.2

a)
P(x <99.6)

z = (x - mean)/s
= (99.6 - 100)/0.2
= 2
P(x <99.6)= P(z <2) = 0.0228

b)
P(99.9 <x < 100.1)
= P((99.9-100)/0.2 < z < ( 100.1 -100)/0.2)
= P(-0.5 < z < 0.5)

P(99.9 <x < 100.1) = P(-0.5 < z < 0.5) = 0.3829

c)
z value at 90% = 1.28

z= ( x -mean)/s
1.28 = ( x - 100)/0.2
x = 100.256

d)
z value at 5% = -1.65 , x = 99.8

z= ( x -mean)/s
-1.65 = ( 99.8 - mean)/0.2
mean = 100.13

Related Questions

Navigate

• Browse (All)
• Browse
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.