Question

The volume of cracker boxes are normally distributed, with a mean of 7 oz and a standard deviation of 0.2 oz. If a box is selected at random, What is the probability of getting an amount between that is at least 7.5 oz

Answer 1

Answer:

0.0062 = 0.62% probability of getting an amount between that is at least 7.5 oz.

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 7 oz and a standard deviation of 0.2 oz.

This means that $\mu = 7, \sigma = 0.2$

What is the probability of getting an amount between that is at least 7.5 oz?

This is 1 subtracted by the pvalue of Z when X = 7.5. So

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

$Z = \frac{7.5 - 7}{0.2}$

$Z = 2.5$

$Z = 2.5$ has a pvalue of 0.9938

1 - 0.9938 = 0.0062

0.0062 = 0.62% probability of getting an amount between that is at least 7.5 oz.