91Ó°ÊÓ

The value of the mean

${\mathrm{μ}}_X=0.13\text{ounce}$

The value of standard deviation,

${\mathrm{σ}}_X=0.02\text{ounce}$

The value of Tube weighs,

$x=0.85-\text{ounce}$

If X and Y are independent,

The Property mean:

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

The Property variance:

${\mathrm{σ}}_{aX+bY}^2=a^2{\mathrm{μ}}_X^2+b^2{\mathrm{μ}}_Y^2$

Now,

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

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

We know that

The square root of variance is standard deviation:

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

Find the z - score:

$\begin{array}{rcl}z& =& \frac{x-\mathrm{μ}}{\mathrm{σ}}\\ & =& \frac{0.85-0.78}{0.0490}\\ & ≈& 1.43\end{array}$

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 $0.0764$.