Question

He has a new 0.85-ounce tube of toothpaste that’s supposed to last him the whole trip. The amount of toothpaste Ken squeezes out of the tube each time he brushes varies according to a Normal distribution with mean 0.13 ounces and standard deviation 0.02 ounces.

Answer 1

Using the normal distribution, it is found that there is a 0.8888 = 88.88% probability that his tube last the whole trip.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• For n instances of a normal variable, the mean is $n\mu$ and the standard deviation is $s = \sigma\sqrt{n}$

In this problem:

• For each time he brushes his teeth, the mean is of 0.13 ounces and the standard deviation is of 0.02 ounces, hence $\mu = 0.13, \sigma = 0.2$

• He will brush his teeth 6 times, hence $n = 6, n\mu = 6(0.13) = 0.79, s = 0.2\sqrt{6}$

The probability that his tube last the whole trip is the p-value of Z when X = 0.85, hence:

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

Considering the n instances

$Z = \frac{X - n\mu}{s}$

$Z = \frac{0.85 - 0.79}{0.2\sqrt{6}}$

$Z = 1.22$

$Z = 1.22$ has a p-value of 0.8888.

0.8888 = 88.88% probability that his tube last the whole trip.

A similar problem is given at https://brainly.com/question/24663213