Question

Sugar packaged by a certain machine has a mean weight of 5 lb and a standard deviation of 0.03 lb. For what values of c can the manufacturer of the machinery claim that the sugar packaged by this machine has a weight between 5 − c and 5 + c lb with probability at least 91%?

Answer 1

That would depend on how the weights are distributed. Can we assume normal?

Answer 2

@SithsAndGiggles I think so, nothing else is said for this problem

Answer 3

Okay, you are asked to find \(c\) such that, if \(W\) denotes the weight of a package of sugar,
\[P(5-c<W<5+c)\le0.91\]
Since \(W\) is normally distributed, transform it into a *standard* normally distributed variable, \(Z\), using the formula:
\[Z=\frac{W-\mu}{\sigma}~~\iff~~W=Z\sigma+\mu\]
where \(\mu\) and \(\sigma\) are the mean and std. deviation of \(W\)'s distribution.
\[P(5-c<0.03Z+5<5+c)\le0.91\\
P\left(-\frac{c}{0.03}<Z<\frac{c}{0.03}\right)\le0.91\\
P\left(-\frac{c}{0.03}<Z<\frac{c}{0.03}\right)\le0.91\]
Due to the symmetry of the normal curve, you can write this as
\[\begin{align*}P\left(0<Z<\frac{c}{0.03}\right)&\le0.455\\
P\left(Z<\frac{c}{0.03}\right)-P(Z<0)&\le0.455\\
P\left(Z<\frac{c}{0.03}\right)-0.5&\le0.455\\
P\left(Z<\frac{c}{0.03}\right)&\le0.955
\end{align*}\]
Referring to a \(z\)-table, you get a corresponding \(z\)-value of 1.70
( http://facstaff.unca.edu/dohse/Stat225/Images/Table-z.JPG )

In other words,
\[P\left(Z<\frac{c}{0.03}\right)\le0.955~~\iff~~P\left(Z<1.70\right)\le0.955\]
which means \(1.7=\dfrac{c}{0.03}~~\Rightarrow~~c=0.051\)

Answer 4

Thank you SO much for showing me the steps for this! :)

Answer 5

You're welcome!