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Chapter 5: Q55E (page 237)

The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \({\rm{ \mu = 50}}\).

a. Calculate the approximate probability that between \({\rm{35 and 70 }}\)tickets are given out on a particular day.

b. Calculate the approximate probability that the total number of tickets given out during a \({\rm{5 - }}\)day week is between \({\rm{225 and 275}}\)

c. Use software to obtain the exact probabilities in (a) and (b) and compare to their approximations.

Short Answer

Expert verified

a. \({\rm{P(35\poundsX\pounds70) = 0}}{\rm{.9838 = 98}}{\rm{.38\% }}\)

b. \({\rm{P(225\poundsX\pounds275) = 0}}{\rm{.8858 = 88}}{\rm{.58\% }}\)

c. \(\begin{array}{*{20}{c}}{{\rm{P(35\poundsX\pounds70) = 0}}{\rm{.986248 = 98}}{\rm{.6248\% }}}\\{{\rm{P(225\poundsX\pounds275) = 0}}{\rm{.893368 = 89}}{\rm{.3368\% }}}\end{array}\)

Step by step solution

01

Definition of probability

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Calculating the approximate probability that between \({\rm{35 and 70}}\)tickets are given out on a particular day.

Given: \({\rm{\mu = 50}}\)

The rule of thumb for a Poisson distribution is that if \({\rm{\mu > 20}}\), the normal distribution should be used to approximate the Poisson distribution.

The Poisson distribution's variance is equal to the Poisson distribution's mean. The square root of the variance is the standard deviation.

\({\rm{\sigma = }}\sqrt {{{\rm{\sigma }}^{\rm{2}}}} {\rm{ = }}\sqrt {\rm{\mu }} {\rm{ = }}\sqrt {{\rm{50}}} \)

The z-value is calculated by subtracting the sample mean from the population mean and dividing by the standard deviation:

\(\begin{array}{*{20}{c}}{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{34}}{\rm{.5 - 50}}}}{{\sqrt {{\rm{50}}} }}{\rm{\gg - 2}}{\rm{.19}}}\\{{\rm{z = }}\frac{{{\rm{\bar x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{70}}{\rm{.5 - 50}}}}{{\sqrt {{\rm{50}}} }}{\rm{\gg 2}}{\rm{.90}}}\end{array}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probability to the left of z-scores).

\(\begin{array}{*{20}{c}}{{\rm{P(35\poundsX\pounds70) = P( - 2}}{\rm{.19 < Z < 2}}{\rm{.90)}}}\\{{\rm{ = P(Z < 2}}{\rm{.90) - P(Z < - 2}}{\rm{.19)}}}\\{{\rm{ = 0}}{\rm{.9981 - 0}}{\rm{.0143}}}\\{{\rm{ = 0}}{\rm{.9838}}}\\{{\rm{ = 98}}{\rm{.38\% }}}\end{array}\)

03

Step 3. Calculating the approximate probability that the total number of tickets given out during a \({\rm{5 - }}\)day week is between \({\rm{225 and 275}}\)

Given:

\({\rm{\mu = 50}}\)

\({\rm{n = 15}}\)

The rule of thumb for a Poisson distribution is that if \({\rm{\mu > 20}}\), the normal distribution should be used to approximate the Poisson distribution.

The sum of five Poisson distributions has a mean that is five times the mean:

\({{\rm{\mu }}_{\rm{X}}}{\rm{ = 5 \times \mu = 5 \times 50 = 250}}\)

The Poisson distribution's variance is equal to the Poisson distribution's mean. The square root of the variance is the standard deviation.

\({{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{{\rm{\sigma }}^{\rm{2}}}} {\rm{ = }}\sqrt {\rm{\mu }} {\rm{ = }}\sqrt {{\rm{250}}} \)

The \({\rm{z}}\)-value is the sample mean divided by the standard deviation of the population mean:

\(\begin{array}{*{20}{c}}{{\rm{z = }}\frac{{{\rm{x - }}{{\rm{\mu }}_{\rm{X}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}{\rm{ = }}\frac{{{\rm{225 - 250}}}}{{\sqrt {{\rm{250}}} }}{\rm{\gg - 1}}{\rm{.58}}}\\{{\rm{z = }}\frac{{{\rm{x - }}{{\rm{\mu }}_{\rm{X}}}}}{{{{\rm{\sigma }}_{\rm{X}}}}}{\rm{ = }}\frac{{{\rm{275 - 250}}}}{{\sqrt {{\rm{250}}} }}{\rm{\gg 1}}{\rm{.58}}}\end{array}\)

Using the normal probability table in the appendix, calculate the corresponding probability (which contains the probability to the left of \({\rm{z}}\)-scores).

\(\begin{array}{*{20}{c}}{{\rm{P(225\poundsX\pounds275) = P( - 1}}{\rm{.58 < Z < 1}}{\rm{.58) = P(Z < 1}}{\rm{.58) - P(Z < - 1}}{\rm{.58)}}}\\{{\rm{ = 0}}{\rm{.9429 - 0}}{\rm{.0571 = 0}}{\rm{.8858 = 88}}{\rm{.58\% }}}\end{array}\)

04

calculating the exact probabilities in (a) and (b) and compare to their approximations.

Poisson probability formula:

\({\rm{P(X = k) = }}\frac{{{{\rm{\mu }}^{\rm{k}}}{{\rm{e}}^{{\rm{ - \mu }}}}}}{{{\rm{k!}}}}\)

Calculate the following probability (using technology):

\({\rm{P(35\poundsX\pounds70) = }}\mathop {\rm{{\aa}}}\limits_{{\rm{k = 35}}}^{{\rm{70}}} {\rm{P(X = k) = }}\mathop {\rm{{\aa}}}\limits_{{\rm{k = 35}}}^{{\rm{70}}} \frac{{{\rm{5}}{{\rm{0}}^{\rm{k}}}{{\rm{e}}^{{\rm{ - 50}}}}}}{{{\rm{k!}}}}{\rm{\gg 0}}{\rm{.986248 = 98}}{\rm{.6248\% }}\)

The sum of five Poisson distributions has a mean that is five times the mean:

\({{\rm{\mu }}_{\rm{X}}}{\rm{ = 5 \times \mu = 5 \times 50 = 250}}\)

Calculate the following probability (using technology):

\({\rm{P(225\poundsX\pounds275) = }}\mathop {\rm{{\aa}}}\limits_{{\rm{k = 225}}}^{{\rm{275}}} {\rm{P(X = k) = }}\mathop {\rm{{\aa}}}\limits_{{\rm{k = 225}}}^{{\rm{275}}} \frac{{{\rm{25}}{{\rm{0}}^{\rm{k}}}{{\rm{e}}^{{\rm{ - 250}}}}}}{{{\rm{k!}}}}{\rm{\gg 0}}{\rm{.893368 = 89}}{\rm{.3368\% }}\)

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Most popular questions from this chapter

Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component \({\rm{2}}\) or component \({\rm{3}}\)functions. Let \({{\rm{X}}_{{\rm{1,}}}}{{\rm{X}}_{\rm{2}}}\), and \({{\rm{X}}_{\rm{3}}}\) denote the lifetimes of components \({\rm{1}}\), \({\rm{2}}\), and \({\rm{3}}\), respectively. Suppose the \({{\rm{X}}_{\rm{i}}}\) 鈥檚 are independent of one another and each \({{\rm{X}}_{\rm{i}}}\) has an exponential distribution with parameter \({\rm{\lambda }}\).

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a. Compute the covariance for \({\rm{X}}\) and \({\rm{Y}}\).

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b. For a random sample of \({\rm{50}}\) such pairs, what is the (approximate) probability that the sample mean courtship time is between \({\rm{100}}\) min and \({\rm{125}}\) min?

c. For a random sample of \({\rm{50}}\) such pairs, what is the (approximate) probability that the total courtship time exceeds \({\rm{150}}\) hrs.?

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