question 6 of 10\nthe weights of 9 - ounce bags of a particular brand of potato chips can be modeled by a…
question 6 of 10\nthe 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.\nwhat percent of 9 - ounce bags of this brand of potato chips weigh between 9 and 9.1 ounces?\nround your answer to 4 decimal places and then convert to a percentage.
Answer
Explanation:
Step1: Calculate z - scores
The formula for the z - score is $z=\frac{x-\mu}{\sigma}$. For $x = 9$, $z_1=\frac{9 - 9.12}{0.05}=\frac{- 0.12}{0.05}=-2.4$. For $x = 9.1$, $z_2=\frac{9.1 - 9.12}{0.05}=\frac{-0.02}{0.05}=-0.4$.
Step2: Find probabilities from z - table
We use the standard normal distribution table. The probability corresponding to $z_1=-2.4$ is $P(Z < - 2.4)=0.0082$. The probability corresponding to $z_2=-0.4$ is $P(Z < - 0.4)=0.3446$.
Step3: Calculate the probability between the two z - scores
$P(-2.4<Z<-0.4)=P(Z < - 0.4)-P(Z < - 2.4)$. $P(-2.4<Z<-0.4)=0.3446 - 0.0082=0.3364$.
Step4: Convert to percentage
To convert to a percentage, we multiply by 100. So the percentage is $0.3364\times100 = 33.6400%$.
Answer:
$33.6400%$