Question

A machine fills sugar boxes in such a way that the weights (in grams) are normally distributed with a mean of 2260 g and a standard deviation of 20 g. Another machine checks the weights and rejects packages in the bottom 1% of weights and the top 1% of weights. Find the the minimum and maximum acceptable weights. Answer using whole numbers and enter the bottom 1% value in blank 1 and the top 1% in blank 2.

Answer 1

Answer:

Minimum acceptable weight: 2213g.

Maximum acceptable weight: 2307g.

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:

$\mu = 2260, \sigma = 20$

Find the the minimum and maximum acceptable weights.

Minimum:

Bottom 1%, which is the value of X when Z has a pvalue of 0.01. So X when Z = -2.33.

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

$-2.33 = \frac{X - 2260}{20}$

$X - 2260 = -2.33*20$

$X = 2213$

The minimum acceptable weight is 2213g.

Maximum:

Top 1%, which is the value of X when Z has a pvalue of 1 - 0.01 = 0.99. So X when Z = 2.33.

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

$2.33 = \frac{X - 2260}{20}$

$X - 2260 = 2.33*20$

$X = 2307$

The maximum acceptable weight is 2307g.