Toothpaste Ken is traveling for his business. 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. If Ken brushes his teeth six times during the trip, what's the probability that he'll use all the toothpaste in the tube? Follow the four-step process.

Given in the question that Toothpaste Ken is traveling for his business. 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. We have to find the probability that he'll use all the toothpaste in the tube.

We need to calculate the probability that he will use the entire bottle of toothpaste. Let's make a variable $T=6X$that represents the total amount of toothpaste he uses when brushing his teeth six times. Our objective is to locate $P(T\le 0.85)$

Mean and standard deviation of $T$,

localid="1649859947223" $\begin{array}{r}{U}_T=0.78\\ {\sigma }_T=0.12\end{array}$

We need to calculate the probability that he will use the entire bottle of toothpaste. That is $P(T\le 0.85)$

Let's standardize

localid="1649859952450" $$\begin{gathered} T=0.85 \\ z=\frac{x-{\mu }_T}{{\sigma }_T} \\ =\frac{0.85-0.78}{0.12} \\ \approx 0.58 \end{gathered}$$

Using the table of normal probability as a guide,

localid="1649859959103" $$\begin{gathered} P(T\le 0.85)=P(z\le 0.58) \\ =0.7190 \end{gathered}$$

There is approximately $71.9\%$chance that he will use all of the toothpaste in the tube