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Let $X=$the number that operate for an hour without failure.

$X$is a binomial random variable.

Number of trials $\left(n\right)=350$
Probability of success $\left(p\right)=0.9990$
To calculate the mean, variance and standard deviation of$X$are:
$E\left(X\right)=np$
$V\left(X\right)=np(1-p)$
$SD\left(X\right)=\sqrt{V\left(X\right)}$

Then,

$X~(350,0.999)$

$P(X=x)=\left(\begin{array}{l}350\\ x\end{array}\right)0.{999}^x(1-0.999{)}^{350-x}$.

The mean and standard deviation of $X$to interpret the value in context.

The mean and standard deviation of $X$is:

$$\begin{gathered} E\left(X\right)=350脳0.999 \\ =349.65 \end{gathered}$$
$$\begin{gathered} V\left(X\right)=350脳0.999(1-0.999) \\ =0.3496 \end{gathered}$$
$$\begin{gathered} SD\left(X\right)=\sqrt{0.3496} \\ =0.5913 \end{gathered}$$

Two engines failed the test. And the model of engine is less reliable. Compute the value of$P(X鈮�348).$

The probability that $X$less than or equal to$348$ is:
$$\begin{gathered} P(X鈮�348)=1-P(X=349)-P(X=350) \\ =1鈭�\left(\begin{array}{l}350\\ 349\end{array}\right){0.999}^{349}(1鈭�0.999{)}^{350鈭�349}鈭�\left(\begin{array}{l}350\\ 350\end{array}\right){0.999}^{350}(1鈭�0.999{)}^{350鈭�350} \end{gathered}$$
$=1-0.24684-0.70456$
$=0.0486$

localid="1650029791570" $$\begin{gathered} P(X鈮�348)鈮�0.0486 \\ =4.86\% \end{gathered}$$

Since the probability is less than $5\%$.