Question

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 is independent, and can be modeled by a Normal distribution with mean $0.13$ ounce and standard deviation $0.02$ ounce. If Ken brushes his teeth six times on a randomly selected trip, what’s the probability that he’ll use all the toothpaste in the tube?

Short answer

Ken's chances of finishing the tube of toothpaste are $0.0764$.

Step-by-step solution

Given information

The value of the mean

${\mu }_X=0.13\text{ounce}$

The value of standard deviation,

${\sigma }_X=0.02\text{ounce}$

The value of Tube weighs,

$x=0.85-\text{ounce}$

Calculation

If X and Y are independent,

The Property mean:

${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

The Property variance:

${\sigma }_{aX+bY}^2=a^2{\mu }_X^2+b^2{\mu }_Y^2$

Now,

Brushing six times' mean is the sum of brushing six times' mean:

$\begin{array}{rcl}{\mu }_{X_1+X_2+X_3+X_4+X_5+X_6}& =& {\mu }_{X_1}+{\mu }_{X_2}+{\mu }_{X_3}+{\mu }_{X_4}++{\mu }_{X_3}+{\mu }_{X_6}\\ & =& 6{\mu }_X\\ & =& 6(0.13)\\ & =& 0.78\text{ounce}\end{array}$

We know that

The square root of variance is standard deviation:

$\begin{array}{rcl}{\sigma }_{X_1+X_2+X_3+X_4+X_5+X_6}& =& \sqrt{{\sigma }_{X_1}^2+{\sigma }_{X_2}^2+{\sigma }_{X_3}^2+{\sigma }_{X_4}^2+{\sigma }_{X_5}^2+{\sigma }_{X_6}^2}\\ & =& \sqrt{6(0.02{)}^2}\\ & \approx & 0.0490\text{ounce}\end{array}$

Find the z - score:

$\begin{array}{rcl}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{0.85-0.78}{0.0490}\\ & \approx & 1.43\end{array}$

To find the relevant probability, use Table - A:

$\begin{array}{rcl}P(X& >& 0.85)=\&P(z>1.43)\\ & =& 1-P(z<1.43)\\ & =& 1-0.9236\\ & =& 0.0764\end{array}$

Thus,

Ken's chances of using the entire tube of toothpaste are little to none $0.0764$.