Question

Math 011- MAC26 Yes, this does look very similar to MAC Assignment 24, but note that now instead of considering the weights of individual boxes, you are considering the average weight of six randomly selected boxes The labels on boxes of Cheesy Poofs say that the box contains 16 ounces of Cheesy Poofs, but that doesn't guarantee that each box contains exactly 16.00 ounces. The machine that fills the boxes doesn't always put out the exact same amount. The amount in each box is Normally distributed, and has a standard deviation is.40 ounces. To help assure that most boxes contain as much or more than the label says, the machine is set so that the mean will be 16.28 ounces. If you purchase 6 boxes of Cheesy Poofs, find the probability that the average amount in your six boxes is... a).less than 16.00 ounces b) .. between 16.20 and 16.30 ounces? b) c).more than 16.50 ounces? d) between 16.20 and 16.26 ounces? d) e) 99.9% of the time, the average amount in six boxes of Cheesy Poofs will be at least

Answer 1

Given $\mu=16.28, \ \sigma=0.4, \ n=6$ .

By using central limit theorem:

$\\\mu_{\bar{x}}=\mu=16.28 \\ \sigma_{\bar{x}}=\sigma/\sqrt{n}=0.4/\sqrt{6}=0.1633$

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a)

$P(\bar{X}<16.00)=P\left ( z<\frac{16.00-16.28}{0.1633} \right )=P(z<-1.71)={\color{Red} 0.0436}$

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b)

$\\P(16.20<\bar{X}<16.30)=P\left (\frac{16.20-16.28}{0.1633}< z<\frac{16.30-16.28}{0.1633} \right ) \\\\=P(-0.49<z<0.12)=0.5478-0.3121={\color{Red} 0.2357}$

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c)

$\\P(\bar{X}>16.50)=P\left ( z>\frac{16.50-16.28}{0.1633} \right )\\\\=P(z>1.35)=1-0.9115={\color{Red} 0.0885}$

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d)

$\\P(16.20<\bar{X}<16.26)=P\left (\frac{16.20-16.28}{0.1633}< z<\frac{16.26-16.28}{0.1633} \right ) \\\\=P(-0.49<z<-0.12)=0.4522-0.3121={\color{Red} 0.1401}$

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e) Let 99.9% of the time, the average amount in six boxes of Cheesy Poofs will be at least x ounces. This means 100-99.9=0.1% or 0.001 proportion of Cheesy Poofs will be below x ounces.

From z table, z score corresponding to area 0.0010 is -3.09.

$z=\frac{x-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\Rightarrow x=\mu_{\bar{x}}+z\sigma_{\bar{x}}=16.28+(-3.09)(0.1633) ={\color{Red} 16.78 \ ounces}$