Question

1. potato chips refer to exercise 42. use the 68 - 95 - 99.7 rule to answer the following questions. show your work! (a) between what weights do the middle 68% of bags fall? (b) what percent of bags weigh less than 9.02 ounces? (c) what percent of 9 - ounce bags of this brand of potato chips weigh between 8.97 and 9.17 ounces? (d) a bag that weighs 9.07 ounces is at what percentile in this distribution?

Explanation:

Step1: Recall 68 - 95 - 99.7 rule

For a normal distribution, middle 68% is within 1 - standard - deviation of mean.

Step2: Answer (a)

Let mean be $\mu$ and standard - deviation be $\sigma$. The middle 68% is between $\mu-\sigma$ and $\mu + \sigma$.

Step3: Answer (b)

Need to know $\mu$ and $\sigma$ to standardize 9.02 and find the proportion.

Step4: Answer (c)

Standardize 8.97 and 9.17 to $z_1=\frac{8.97 - \mu}{\sigma}$ and $z_2=\frac{9.17 - \mu}{\sigma}$, then find the proportion between them.

Step5: Answer (d)

Standardize 9.07 to $z=\frac{9.07 - \mu}{\sigma}$, then find the percentile from the standard normal table.

Since no mean and standard - deviation are given, we assume a general normal - distribution approach. But without specific values, we can't give numerical answers. If we assume mean $\mu = 9$ and standard - deviation $\sigma=0.05$ (for illustration purposes only):

Step2 (revised for (a))

$\mu-\sigma=9 - 0.05=8.95$ and $\mu+\sigma=9 + 0.05 = 9.05$.

Step3 (revised for (b))

$z=\frac{9.02 - 9}{0.05}=0.4$. Looking up in the standard - normal table, the proportion is 0.6554 or 65.54%.

Step4 (revised for (c))

$z_1=\frac{8.97 - 9}{0.05}=-0.6$ and $z_2=\frac{9.17 - 9}{0.05}=3.4$. Proportion between them is $P(-0.6

Step5 (revised for (d))

$z=\frac{9.07 - 9}{0.05}=1.4$. Looking up in the standard - normal table, the percentile is 0.9192 or 91.92%.

Answer:

(a) Assuming $\mu = 9$ and $\sigma = 0.05$, between 8.95 and 9.05 ounces.
(b) Assuming $\mu = 9$ and $\sigma = 0.05$, 65.54%.
(c) Assuming $\mu = 9$ and $\sigma = 0.05$, 72.54%.
(d) Assuming $\mu = 9$ and $\sigma = 0.05$, 91.92%.