Question

A machine in the JLM packaging plant is used to fill individual-serving coffee brewing containers with 9.4 g of coffee. The average actual weight of the coffee in each container is found to be 9.50 g with a standard deviation of 0.05 g. From past experience, it is known that the weights are normally distributed. If a batch run consists of 210000 containers, how many containers in a batch can be expected to weigh less than 9.4 g? Write an integer as your answer. Do not round any intermediate calculations. The number of containers in a batch that can be expected to weigh less than 9.4 g is

Answer 1

Answer:

The number of containers in a batch that can be expected to weigh less than 9.4 g is 4788

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 9.5, \sigma = 0.05$

If a batch run consists of 210000 containers, how many containers in a batch can be expected to weigh less than 9.4 g?

The first step is finding the proportion of containers in a batch that are expected to weigh less than 9.4g. This is the pvalue of Z when X = 9.4. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{9.4 - 9.5}{0.05}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228.

So 2.28% of the containers in a batch can be expected to weigh less than 9.4g.

Out of 210,000 containers

0.0228*210000 = 4,788

The number of containers in a batch that can be expected to weigh less than 9.4 g is 4788