Chapter 6.2, Problem 64E

To determine

To calculate; the probability that all the paste in tube is being used.

Answer to Problem 64E

  $\text{0}\text{.0764}$

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_X=0.13\\ {\text{σ}}_X\text{ = 0}\text{.02}\end{array}$

  $\begin{array}{l}{\text{μ}}_{\text{aX + bY}}{\text{ = a(μ}}_{\text{X}})+b({\text{μ}}_Y)\\ {\text{σ}}^{\text{2}}{}_{\text{aX + bY}}{\text{ = a}}^{\text{2}}{\text{(σ}}^2{}_{\text{X}})+{\text{b}}^{\text{2}}{\text{(σ}}^2{}_{\text{Y}})\end{array}$

  $\begin{array}{l}{\mu }_{X_1+X_2+X_3+X_4+X_5+X_6}={\mu }_{X_1}+{\mu }_{X_2}+{\mu }_{X_3}+{\mu }_{X_4}+{\mu }_{X_5}+{\mu }_{X_6}=6{\mu }_{X_1}=6(0.13)=0.78\\ {\text{σ}}_{{}_{X_1+X_2+X_3+X_4+X_5+X_6}}\text{= }\sqrt{{\text{σ}}^2{}_{X_1}+{\text{σ}}^2{}_{X_2}+{\text{σ}}^2{}_{X_3}+{\text{σ}}^2{}_{X_4}+{\text{σ}}^2{}_{X_5}{\text{+σ}}^2{}_{X_6}}=\sqrt{6{(0.02)}^2}\approx 0.0490\end{array}$

  $z=\frac{x-\mu }{\sigma }=\frac{(0.85-0.78)}{0.0490}\approx 1.43$

  $\text{P( X > 0}\text{.85) = P(Z < 1}\text{.43) = 1 - P(Z > 1}\text{.43) = 1 - 0}\text{.9236 = 0}\text{.0764}$