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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

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean \({\rm{2}}{\rm{.65 }}\)and standard deviation \({\rm{.85}}\) (suggested in 鈥淢odeling Sediment and Water Column Interactions for Hydrophobic Pollutants,鈥� Water Research, \({\rm{1984: 1164 - 1174 }}\)).

a. If a random sample of \({\rm{25}}\)specimens is selected, what is the probability that the sample average sediment density is at most \({\rm{3}}{\rm{.00 }}\)? Between \({\rm{2}}{\rm{.65 }}\)and \({\rm{3}}{\rm{.00 }}\)?

b. How large a sample size would be required to ensure that the first probability in part (a) is at least \({\rm{.99}}\)?

A certain market has both an express checkout line and a superexpress checkout line. Let \({{\rm{X}}_{\rm{1}}}\) denote the number of customers in line at the express checkout at a particular time of day, and let \({{\rm{X}}_{\rm{2}}}\) denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of \({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\) is as given in the accompanying table

a. What is \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = 1}}{{\rm{X}}_{\rm{2}}} = 1)\), that is, the probability that there is exactly one customer in each line?

b. What is \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{X}}_{\rm{2}}})\),that is, the probability that the numbers of customers in the two lines are identical? c. Let A denote the event that there are at least two more customers in one line than in the other line. Express A in terms of \({{\rm{X}}_{\rm{1}}}\) and \({{\rm{X}}_{\rm{2}}}\), and calculate the probability of this event.

d. What is the probability that the total number of customers in the two lines is exactly four? At least four?

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let \({\rm{X = }}\) the number of points earned on the first part and \({\rm{Y = }}\) the number of points earned on the second part. Suppose that the joint pmf of \({\rm{X}}\) and \({\rm{Y}}\) is given in the accompanying table.

a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score\({\rm{E(X + Y)}}\)?

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