Question

the weights of 9 - ounce bags of a particular brand of potato chips can be modeled by a normal distribution with mean (mu = 9.12) ounces and standard deviation (sigma=0.05) ounce. what proportion of 9 - ounce bags of this brand of potato chips weigh more than 9.2 ounces? round your answer to 4 decimal places.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 9.2$, $\mu=9.12$, and $\sigma = 0.05$. So, $z=\frac{9.2 - 9.12}{0.05}$.
$z=\frac{0.08}{0.05}=1.6$

Step2: Find the proportion

We want to find $P(X>9.2)$, which is equivalent to $1 - P(X\leq9.2)$ in a normal - distribution. Looking up the z - score of $z = 1.6$ in the standard normal distribution table, we find that $P(Z\leq1.6)=0.9452$. So, $P(X>9.2)=1 - 0.9452 = 0.0548$.

Answer:

$0.0548$