Question

In an effort to reduce plastics, a shampoo bottle filling process is using machines to fill reusable bottles labeled 200 ml. The amount of shampoo filled in a bottle is normally distributed with the mean 205 ml and standard deviation 2 ml.

a. A bottle filled with more than 210 ml of shampoo will overflow. Suppose you need to ensure that no more than 0.15% of the bottles will overflow in the filling process. Assume that you can calibrate the filling machine to change the mean amount of shampoo filled to a bottle, keeping the standard deviation constant. What should the new mean amount of shampoo filled to a bottle be to ensure that no more than 0.15% of the bottles will overflow in the filling process?

b. Once filled with shampoo, these bottles are packaged into sets of 16 bottles. What percentage of these packages would have a mean volume exceeding 204 ml? Show your work clearly and justify your reasoning.

Answer 1

X: The amount of shampoo filled in a bottle

X follows a Normal distribution with mean = 205 and standard deviation : = 2

a.

Let be the new mean amount of shampoo filled to a bottle be to ensure that no more than 0.15% of the bottles will overflow in the filling process

A bottle filled with more than 210 ml of shampoo will overflow

i.e

P(X > 210) = 1 - P(X 210) = 0.15

P(X 210) = 1-0.15 = 0.85

Z₁ : Z-score for 210

P(X 210) =P(Z Z₁) = 0.85

i.eZ₁ = (210 - ) / = (210-)/2

i.e = 210 - 2Z₁

From Standard Normal tables find Z₁ such that P(Z Z₁) = 0.85

Z₁ = 1.035

= 210 -2 Z₁ = 210-2x1.035 = 207.93

The new mean to be 207.93

b.

If X follows a Normal Distribution with mean and standard deviation , then the Sampling Distribution of sample mean follows a normal distribution with mean and standard deviation (n : sample size)

Given,

bottles are packaged into sets of 16 bottles; i.e sample size n : 16

mean volume i.e follows normal with mean 205 and standard deviation

Probability that these packages would have a mean volume exceeding 204 ml = P( > 204)

P( > 204) = 1-P(204)

P(204)

Z-score for 204 = (204 - 205)/0.5 = -2

P(Z-2) = 0.0228

P(204) = P(Z-2) = 0.0228

P( > 204) =1-P(204) = 1- 0.0228 = 0.9772

Probability that these packages would have a mean volume exceeding 204 ml = P( > 204) = 0.9772

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Not sure whether to use new mean or old mean:

In case new mean to be used then,

mean volume i.e follows normal with mean 207.93 and standard deviation

Probability that these packages would have a mean volume exceeding 204 ml = P( > 204)

P( > 204) = 1-P(204)

P(204)

Z-score for 204 = (204 - 207.93)/0.5 = -7.86

P(Z-7.86) = 0.000

P(204) = P(Z-7.86) = 0.0000

P( > 204) =1-P(204) = 1- 0.0000 = 1.000

Probability that these packages would have a mean volume exceeding 204 ml = P( > 204) = 1.0000