Question

Centerpieces are normally distributed with mean of 30 ounces and standard deviation of 2 ounces. The weights of the shipping boxes are normally distributed with mean of 12 ounces and standard deviation of 1 ounce. Suppose that centerpieces are chosen at random and packed into randomly chosen shipping boxes.If a package exceeds 45 ounces, an additional charge is incurred. What is the proportion of packages will incur such a charge

Answer 1

Answer:

0.0901 = 9.01% of packages will incur such a charge

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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.

Sum of normal variables:

When we add normal variables, the mean will be the sum of the means, while the standard deviation is the square root of the sum of the variances.

Centerpieces are normally distributed with mean of 30 ounces and standard deviation of 2 ounces.

This means that $\mu_C = 30, \sigma_C = 2$

The weights of the shipping boxes are normally distributed with mean of 12 ounces and standard deviation of 1 ounce.

This means that $\mu_S = 12, \sigma_S = 1$

Suppose that centerpieces are chosen at random and packed into randomly chosen shipping boxes.

The shipment is given by the centerpice plus the shipping box. So the mean and the standard deviation of the weight are given by:

$\mu = \mu_C + \mu_S = 30 + 12 = 42$

$\sigma = \sqrt{\sigma_C^2 + \sigma_S^2} = \sqrt{2^2 + 1^2} = \sqrt{5} = 2.236$

If a package exceeds 45 ounces, an additional charge is incurred. What is the proportion of packages will incur such a charge

This proportion is 1 subtracted by the pvalue of Z when X = 45. So

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

$Z = \frac{45 - 42}{2.236}$

$Z = 1.34$

$Z = 1.34$ has a pvalue of 0.9099

1 - 0.9099 = 0.0901

0.0901 = 9.01% of packages will incur such a charge