I am having trouble understanding this particular problem:
A orange juice company fills bottles of orange juice by setting a timer on a filling machine. It has been found that the distribution of the amount of orange juice the machine puts into a bottle is normal with standard deviation of 1 ml. If the company wants 88.8% of all its bottles of orange juice to have at least 375 ml as printed on the container, what mean amount should the filling machine be set to put in each can?
How do I go abouts answering this? I figured that I should use the z score to do some algebra to work out the mean but I'm also not sure where the z score value comes in to this
$\endgroup$1 Answer
$\begingroup$Hint: You can standardize the random variable of the amount of oranges/orange juice $(X)$ to obtain a variable which is nomally distributed as $Z\sim \mathcal N(0,1)$:
$Z=\frac{X-\mu}{\sigma}$
You want $P(X\geq 375)$. Here you can use the converse probablity.
$P(X\geq 375)=1-P(X\leq 375)=1-\Phi\left( \frac{x-\mu}{\sigma} \right)=1-\Phi\left( \frac{375-\mu}{1} \right)=0.888$
$\Phi(z)$ is the cdf of the standard normal distribution. $\mu$ is the mean amount of the oranges. And $\sigma$ is the standard deviation of the amount of the oranges.
Solve the equation for $\mu$
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