91Ó°ÊÓ

Probability: Probability means possibility. It is a branch of mathematics which deals with the occurrence of a random event.

(a)

Considering the given information:

It is assumed that X denotes the number of nonconforming steel shafts that can be reworked among the produced steel shafts. Then we can say that X follows a binomial distribution with the following parameters:

\(\begin{array}{l}{\rm{n = 200}}\\{\rm{p = 0}}{\rm{.1}}\end{array}\)

The mean and standard deviation of X are then given as follows:

\(\begin{array}{c}{\rm{\mu = n \times p}}\\{\rm{ = 200(0}}{\rm{.1)}}\\{\rm{ = 20}}\\{\rm{\sigma = }}\sqrt {{\rm{n \times p \times (1 - p)}}} \\{\rm{ = 4}}{\rm{.24}}\end{array}\)

As a result, we can say that X is roughly distributed as N. (20,4.24)

We now compute \(P(X \le 30)\). It becomes \(P(X \le 30.5)\) due to continuity correlation. Standardization results in:

\(X \le 30.5\)

if and only if

\(\begin{array}{c}\frac{{X - 20}}{{4.24}} \le \frac{{30.5 - 20}}{{4.24}}\\\frac{{X - 20}}{{4.24}} \le \frac{{10.5}}{{4.24}}\\Z \le 2.48\end{array}\)

Thus

\(\begin{array}{c}P(X \le 30) = P(Z \le 2.48)\\ = \phi (2.48)\\P(X \le 30) = 0.9934\end{array}\)

Let Z be a continuous rv with the cdf \(\phi (z)\). Then, for any a and b with a greater than b,

\(\begin{array}{c}P(b \le Z \le a) = \phi (a) - \phi (b)\\P(Z \le a) = \phi (a)\end{array}\)

(b)

Considering the given information:

Because X can only take integer values,

\(P(X < 30) = P(X \le 29)\)

We now compute \(P(X \le 29)\). It becomes \(P(X \le 29.5)\)due to continuity correlation. The benefits of standardisation include:

\(X \le 29.5\)

if and only if

\(\begin{array}{c}\frac{{X - 20}}{{4.24}} \le \frac{{29.5 - 20}}{{4.24}}\\\frac{{X - 20}}{{4.24}} \le \frac{{9.5}}{{4.24}}\\Z \le 2.24\end{array}\)

Thus

\(\begin{array}{c}P(X < 30) = P(X \le 29)\\ = P(Z \le 2.24)\\ = \phi (2.24)\\P(X < 30) = 0.9875\end{array}\)

Therefore, the required value is \({\rm{0}}{\rm{.9875}}\).

(c)

Considering the given information:

We need to figure out\(P(15 \le X \le 25)\). It becomes\(P\left( {14.5 \le {X_{{\rm{normal }}}} \le 25.5} \right)\)The benefits of standardisation include:

\(14.5 \le X \le 25.5\)

if and only if

\(\begin{array}{c}\frac{{14.5 - 20}}{{4.24}} \le \frac{{X - 20}}{{4.24}} \le \frac{{25.5 - 20}}{{4.24}}\\\frac{{ - 5.5}}{{4.24}} \le \frac{{X - 20}}{{4.24}} \le \frac{{5.5}}{{4.24}}\\ - 1.3 \le Z \le 1.3\end{array}\)

Thus

\(\begin{array}{c}P(15 \le X \le 25) = P( - 1.3 \le Z \le 1.3)\\ = \phi (1.3) - \phi ( - 1.3)\\ = 0.9032 - 0.0968\\P(15 \le X \le 25) = 0.8064\end{array}\)

Therefore, the required value is\({\rm{0}}{\rm{.8064}}\).