Question

The weights of bags of peas are normally distributed with a mean of 13.50 ounces and a standard deviation of 1.06 ounces. Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?

Answer 1

Answer:

The most that a bag can weigh and not need to be repackaged is 15.355 ounces.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 13.50, \sigma = 1.06$

Bags in the upper 4% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?

Bags in the upper 4% have a pvalue of 1-0.04 = 0.96. So the most that a bag can weight and not need to be repackaged is the value of X when Z has a pvalue of 0.9599. So this is X when $Z = 1.75$

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

$1.75 = \frac{X - 13.50}{1.06}$

$X - 13.50 = 1.75*1.06$

$X = 15.355$

The most that a bag can weigh and not need to be repackaged is 15.355 ounces.