Question

question 6 of 10 > 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. use the empirical rule to answer the following questions. (a) what percent of bags weigh less than 9.07 ounces? 16 % correct answer (do not round.) (b) what proportion of bags weigh between 8.97 and 9.17 ounces? (enter your answer in decimal form.) 83.85 incorrect answer (do not round.)

Explanation:

Step1: Recall the empirical - rule for normal distribution

The empirical rule states that for a normal distribution: about 68% of the data lies within 1 standard - deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 standard - deviations of the mean ($\mu\pm2\sigma$), and about 99.7% lies within 3 standard - deviations of the mean ($\mu\pm3\sigma$).

Step2: Calculate the z - scores and use the empirical rule for part (a)

Given $\mu = 9.12$ and $\sigma=0.05$. For $x = 9.07$, the z - score is $z=\frac{x - \mu}{\sigma}=\frac{9.07 - 9.12}{0.05}=\frac{-0.05}{0.05}=-1$. Since about 68% of the data lies within $\mu\pm\sigma$ (i.e., between $z=-1$ and $z = 1$), the percentage of data to the left of $z=-1$ is $\frac{100 - 68}{2}=16\%$.

Step3: Calculate the z - scores and use the empirical rule for part (b)

For $x_1 = 8.97$, $z_1=\frac{8.97 - 9.12}{0.05}=\frac{-0.15}{0.05}=-3$. For $x_2 = 9.17$, $z_2=\frac{9.17 - 9.12}{0.05}=\frac{0.05}{0.05}=1$.
The percentage of data within $\mu\pm3\sigma$ is 99.7% and within $\mu\pm1\sigma$ is 68%. The percentage of data between $z=-3$ and $z = 1$ is $\frac{99.7+(68)}{2}=83.85\%$. In decimal form, the proportion is 0.8385.

Answer:

(a) 16
(b) 0.8385