91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability

It is given that \({\rm{X}}\) denotes the number of defective circuit boards among the batch of \({\rm{250}}\) boards. Since after disc players are tested, the long-run percentage of defectives is\({\rm{5\% }}\). Then we can say that \({\rm{X}}\) have a binomial distribution with parameters:

\(\begin{array}{l}{\rm{n = 250}}\\{\rm{p = 0}}{\rm{.05}}\end{array}\)

Now we check the following products:

\(\begin{array}{l}{\rm{np = 250(0}}{\rm{.05) = 12}}{\rm{.5}}\\{\rm{nq = 250(1 - 0}}{\rm{.05) = 237}}{\rm{.5}}\end{array}\)

Since: and

Hence, we can use the normal approximation. Then mean and standard deviation of \({\rm{X}}\)are given as:

\(\begin{array}{l}{\rm{\mu = n \times p = 250(0}}{\rm{.05) = 12}}{\rm{.5}}\\{\rm{\sigma = }}\sqrt {{\rm{n \times p \times (1 - p)}}} {\rm{ = 3}}{\rm{.446}}\end{array}\)

Hence, we can say that \({\rm{X}}\) is approximately distributed as \({\rm{N(12}}{\rm{.5,3}}{\rm{.446)}}\)

(a) Since \({\rm{10\% }}\) of \({\rm{250}}\) is\({\rm{25}}\), Hence the approximate probability that at least \({\rm{10\% }}\) of the boards in the batch are defective can be represented as. Continuity condition and standardizing gives:

\({{\text{X}}_{{\text{normal }}}}{\text{24}}{\text{.5}}\)

if and only if

\(\frac{{{X_{normal{\text{ }}}} - 12.5}}{{3.446}}\frac{{24.5 - 12.5}}{{3.446}}\frac{{{X_{normal{\text{ }}}} - 12.5}}{{3.446}}\frac{{12}}{{3.48}}Z3.48\)

Thus

\(\begin{aligned}P\left( {{X_{bin}}25} \right) &= P(Z3.48) \\&= 1 - f(3.48) \\&= 1 - 0.9997P\left( {{X_{bin}}25} \right) \\&= 0.0003 \\\end{aligned} \)

Proposition: Let \({\rm{Z}}\) be a continuous rv with cdf. Then for any \({\rm{a}}\) and \({\rm{b}}\) with\({\rm{a > b}}\),

\(\begin{aligned}P(b£Z£a) &= f(a) - f(b) \hfill \\P(Z£a) &= f(a) \hfill \\\end{aligned} \)

(b) The approximate probability that there are exactly \({\rm{10}}\) defectives in the batch can be represented as \({\rm{P}}\left( {{{\rm{X}}_{{\rm{bin }}}}{\rm{ = 10}}} \right)\)

Using continuity correlation, it becomes:

\(P\left( {{X_{bin{\text{ }}}} = 10} \right) = P\left( {9.5£{X_{normal}}£10.5} \right)\)

Standardizing gives:

°À(9.5£³Ý£10.5°À)

if and only if

\(\frac{{9.5 - 12.5}}{{3.446}}£\frac{{X - 12.5}}{{3.446}}£\frac{{10.5 - 12.5}}{{3.446}}\frac{{ - 3}}{{3.446}}£\frac{{X - 12.5}}{{3.446}}£\frac{{ - 2}}{{3.446}} - 0.87£Z£- 0.58\)

Thus

\(\begin{aligned}P\left( {{X_{bin{\text{ }}}} = 10} \right) &= P( - 0.87£Z£ - 0.58) \\&= f( - 0.58) - f( - 0.87) \\&= 0.2810 - 0.1922P\left( {{X_{bin{\text{ }}}} = 10} \right) \\&= 0.0888 \\\end{aligned} \)