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The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

1)

The probability that the droplet size is less than \({\rm{1500\mu m}}\)is denoted as\({\rm{P(X < 1500)}}\). Standardizing gives:

\({\rm{X < 1500}}\)

if and only if

\(\frac{{{\rm{X - 1050}}}}{{{\rm{150}}}}{\rm{ < }}\frac{{{\rm{1500 - 1050}}}}{{{\rm{150}}}}\frac{{{\rm{X - 1050}}}}{{{\rm{150}}}}{\rm{ < }}\frac{{{\rm{450}}}}{{{\rm{150}}}}{\rm{Z < 3}}\)

Thus

\({\rm{P(X < 1500) = P(Z < 3)}}\)

Here \({\rm{Z}}\) is a standard normal distribution \({\rm{rv}}\) with \(cdf\phi (z)\). Hence

\(P(X < 1500) = P(Z < 3) = \phi (3)\)

To get\(\phi (3)\), we check Appendix Table \({\rm{A}}{\rm{.3}}\)at the intersection of the row marked .03 and the column marked. \({\rm{00}}\), from there

\(\phi (3) = 0.9987\)

Hence

\({\rm{P(X < 1500) = 0}}{\rm{.9987}}\)

The probability that the droplet size is at least \({\rm{1000\mu m}}\) is denoted as\(P(X \ge 1000)\). Standardizing gives:

\(X \ge 1000\)

if and only if

\(\frac{{X - 1050}}{{150}} \ge \frac{{1000 - 1050}}{{150}}\frac{{X - 1050}}{{150}} \ge \frac{{ - 50}}{{150}}Z \ge - 0.33\)

Thus

\(P(X \ge 1000) = P(Z \ge - 0.33)\)

Here \({\rm{Z}}\)is a standard normal distribution \({\rm{rv}}\) with \(cdf\phi (z)\). Hence

\(P(X \ge 1000) = P(Z \ge - 0.33) = 1 - \phi ( - 0.33)\)

To get\(\phi ( - 0.33)\), we check Appendix Table \({\rm{A}}{\rm{.3}}\), from there

\(\begin{array}{l}{\rm{f( - 0}}{\rm{.33) = 0}}{\rm{.37071 - f( - 0}}{\rm{.33)}}\\{\rm{ = 1 - 0}}{\rm{.3707 = 0}}{\rm{.6293}}\end{array}\)

Hence

\(P(X \ge 1000) = 0.6293\)

Proposition: Let Z be a continuous \({\rm{rv}}\) with \(cdf\phi (z)\). Then for any number a,

\(\begin{array}{l}P(Z > a) = 1 - \phi (a)\\P(Z < a) = \phi (a)\end{array}\)

And for any two numbers \({\rm{a}}\)and \({\rm{b}}\) with\({\rm{a < b}}\),

\(P(a < Z < b) = \phi (b) - \phi (a)\)

Therefore, the probabilities are \(P(X < 1500) = 0.9987\)and \(P(X \ge 1000) = 0.6293\).

2)

(2) The probability that the size of a single droplet is between \({\rm{1000}}\) and \({\rm{1500\mu m}}\)is denoted as\({\rm{P(1000 < X < 1500)}}\).

We have already calculated following probabilities in part (a):

\(\begin{array}{l}{\rm{P(X > 1000) = 0}}{\rm{.629}}\\{\rm{3P(X < 1500) = 0}}{\rm{.9987}}\end{array}\)

Using these we can write:

\(\begin{array}{c}{\rm{P(1000 < X < 1500) = P(X < 1500) - P(X < 1000)}}\\{\rm{ = 0}}{\rm{.9987 - (1 - 0}}{\rm{.6293)P(1000 < X < 1500)}}\\{\rm{ = 0}}{\rm{.628}}\end{array}\)

Therefore, the probability is \({\rm{0}}{\rm{.628}}\)that the size of a single droplet.

3)

(3) The smallest \({\rm{2\% }}\)of droplet sizes (less than \({\rm{2}}\) percentile) denote all the values smaller than \({{\rm{Z}}_{{\rm{0}}{\rm{.98}}}}\)

If we recall, according to the definition, \({{\rm{z}}_{\rm{\alpha }}}\)is the \({\rm{100(1 - \alpha }}{{\rm{)}}^{{\rm{th }}}}\)percentile of the standard normal distribution.

\({{\rm{z}}_{{\rm{0}}{\rm{.98}}}}\)means that area to the left of \({{\rm{z}}_{{\rm{0}}{\rm{.98}}}}\)under standard normal distribution curve is \({\rm{0}}{\rm{.02}}\)

we can also say that:

\(\phi \left( {{z_{0.98}}} \right) = 0.02\)

Where \(\phi (z)\) is the \({\rm{cdf}}\) of standard normal distributed \({\rm{rv}}\) \({\rm{z}}\).

We check Appendix Table \({\rm{A}}{\rm{. 3}}\)to see if \(\phi (z)\)is equal to $0.02$ for any\({\rm{z}}\).

The two closest values to

\(\begin{array}{c}\frac{{{{\rm{X}}_{{\rm{0}}{\rm{.98}}}}{\rm{ - 1050}}}}{{{\rm{150}}}}{\rm{ = }}{{\rm{Z}}_{{\rm{0}}{\rm{.98}}}}\frac{{{{\rm{X}}_{{\rm{0}}{\rm{.98}}}}{\rm{ - 1050}}}}{{{\rm{150}}}}\\{\rm{ = - 2}}{\rm{.054}}{{\rm{X}}_{{\rm{0}}{\rm{.98}}}}{\rm{ = 1050 + (150)( - 2}}{\rm{.054)}}{{\rm{X}}_{{\rm{0}}{\rm{.98}}}}\\{\rm{ = 741}}{\rm{.9}}\end{array}\)

Hence the smallest \({\rm{2\% }}\)of all droplet sizes are smaller than \({\rm{741}}{\rm{.9\mu m 0}}{\rm{.02}}\) are \({\rm{0}}{\rm{.0202}}\)and \({\rm{0}}{\rm{.0197}}\)which belong to \({\rm{z}}\)-values of \({\rm{ - 2}}{\rm{.05}}\)and \({\rm{ - 2}}{\rm{.06}}\)respectively, Hence using interpolation:

\(\begin{array}{c}\frac{{{{\rm{z}}_{{\rm{0}}{\rm{.98}}}}{\rm{ - ( - 2}}{\rm{.06)}}}}{{{\rm{0}}{\rm{.02 - 0}}{\rm{.0197}}}}{\rm{ = }}\frac{{{\rm{ - 2}}{\rm{.05 - ( - 2}}{\rm{.06)}}}}{{{\rm{0}}{\rm{.0202 - 0}}{\rm{.0197}}}}{{\rm{z}}_{{\rm{0}}{\rm{.98}}}}\\{\rm{ = - 2}}{\rm{.054}}\end{array}\)

Let \({{\rm{X}}_{{\rm{0}}{\rm{.98}}}}\)denote the value of \({\rm{rv}}\) corresponding to \({\rm{z}}\)-value of\({{\rm{Z}}_{{\rm{0}}{\rm{.98}}}}\).

Therefore, The smallest \({\rm{2\% }}\)of all droplet sizes are smaller than \({\rm{741}}{\rm{.9\mu m}}{\rm{.}}\)

4)

(4) Let \({\rm{P(E)}}\)denote the probability that if the sizes of five independently selected droplets are measured, exactly two of them exceed \({\rm{1500\mu m}}\).

Let \({\rm{p}}\)denote the probability that size of a randomly selected droplet exceeds\({\rm{1500\mu m}}\). Then as we have calculated in part(a):

\(\begin{array}{c}{\rm{P(X < 1500) = 0}}{\rm{.9987p}}\\{\rm{ = 1 - P(X < 1500)}}\\{\rm{ = 1 - 0}}{\rm{.9987p}}\\{\rm{ = 0}}{\rm{.0013}}\end{array}\)

Then \({\rm{P(E)}}\)can be written in terms \({\rm{of p}}\)as:

\(\begin{array}{c}{\rm{P(E) = C}}_{\rm{2}}^{\rm{5}}{\rm{ \times }}{{\rm{p}}^{\rm{2}}}{\rm{ \times (1 - p}}{{\rm{)}}^{\rm{3}}}{\rm{ = 10 \times (0}}{\rm{.0013}}{{\rm{)}}^{\rm{2}}}{\rm{ \times (0}}{\rm{.9987}}{{\rm{)}}^{\rm{3}}}{\rm{P(E)}}\\{\rm{ = 1}}{\rm{.68 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\end{array}\)

Therefore, the probability \({\rm{1}}{\rm{.68 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\) that exactly two of them exceed \({\rm{1500\mu m}}\).