91Ó°ÊÓ

Given :

Mean,

${\mathrm{μ}}_A=62$seconds

${\mathrm{μ}}_B=62.8$ seconds

Standard deviation,

${\mathrm{σ}}_A=0.8$seconds

${\mathrm{σ}}_B=1$second

Because $\mathrm{A}\mathrm{and}\mathrm{B}$have a Normal distribution, their difference has Normal distribution as well.

The mean of the difference is now equal to the difference in the means of any two variables.

$$\begin{gathered} {\mathrm{μ}}_{A-B}={\mathrm{μ}}_X-{\mathrm{μ}}_Y \\ =62-62.8 \\ =-0.8\mathrm{seconds} \end{gathered}$$

The variance of the difference is equal to the sum of the variances of the random variables when they are independent.

$$\begin{gathered} {{\mathrm{σ}}^2}_{A-B}={{\mathrm{σ}}^2}_A+{{\mathrm{σ}}^2}_B \\ ={(0.8)}^2+{\left(1\right)}^2 \\ =1.64\mathrm{seconds} \end{gathered}$$

The standard deviation is equal to the square root of the variance, as we know.

$$\begin{gathered} {\mathrm{σ}}_{x+y}=\sqrt{{{\mathrm{σ}}^2}_{A-B}} \\ =\sqrt{1.64} \\ ≈\mathbf{1}\mathbf{.}\mathbf{2806}\mathrm{seconds} \end{gathered}$$

Because Andrea loses to Barsha, this implies that $A>B\mathrm{or}(A-B>0)$

We must next consider $x=0$. Calculate the$z-$score, which is equal to

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0-(-0.5)}{1.2806} \\ ≈0.39 \end{gathered}$$

To find the equivalent probability, use the normal probability table in the appendix.

$P(z<0.39)$can be found in the standard normal probability table in the row that starts with $0.3$and the column that starts with $.09$.

$$\begin{gathered} P(A-B>0)=P(z>0.39) \\ =1-P(z<0.39) \\ =1-0.6517 \\ =\mathbf{}\mathbf{0}\mathbf{.}\mathbf{3483} \end{gathered}$$