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Chapter 5: Q87SE (page 245)

Garbage trucks entering a particular waste-management facility are weighed prior to offloading their contents. Let \({\rm{X = }}\)the total processing time for a randomly selected truck at this facility (waiting, weighing, and offloading). The article "Estimating Waste Transfer Station Delays Using GPS" (Waste Mgmt., \({\rm{2008: 1742 - 1750}}\)) suggests the plausibility of a normal distribution with mean \({\rm{13\;min}}\)and standard deviation \({\rm{4\;min}}\)for\({\rm{X}}\). Assume that this is in fact the correct distribution.

a. What is the probability that a single truck's processing time is between \({\rm{12}}\) and \({\rm{15\;min}}\)?

b. Consider a random sample of \({\rm{16}}\) trucks. What is the probability that the sample mean processing time is between \({\rm{12}}\) and\({\rm{15\;min}}\)?

c. Why is the probability in (b) much larger than the probability in (a)?

d. What is the probability that the sample mean processing time for a random sample of \({\rm{16}}\) trucks will be at least\({\rm{20\;min}}\)?

Short Answer

Expert verified

a) The probability is \({\rm{P(12 < X < 15) = 0}}{\rm{.2902 = 29}}{\rm{.02\% }}\).

b) The probability is \({\rm{P(12 < \bar X < 15) = 0}}{\rm{.8185 = 81}}{\rm{.85\% }}\).

c) The sample mean of multiple trucks is less variable than a single truck.

d) The probability is \({\rm{P(\bar X > 20)\gg 0 = 0\% }}\)

Step by step solution

01

Definition

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.

02

Calculating probability

a)

Given: \({\rm{X}}\)has a normal distribution with

\(\begin{array}{l}{\rm{\mu = 13\;min}}\\{\rm{\sigma = 4\;min}}\end{array}\)

The standardized score is calculated by dividing the value\({\rm{x}}\)by the mean and then by the standard deviation.

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{12 - 13}}}}{{\rm{4}}}\\{\rm{\gg - 0}}{\rm{.25}}\\{\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}\\{\rm{ = }}\frac{{{\rm{15 - 13}}}}{{\rm{4}}}\\{\rm{\gg 0}}{\rm{.50}}\end{array}\)

Determine the relevant probability using the appendix's normal probability table (which includes the probability to the left of z-scores:

\(\begin{array}{c}{\rm{P(12 < X < 15) = P( - 0}}{\rm{.25 < Z < 0}}{\rm{.50)}}\\{\rm{ = P(Z < 0}}{\rm{.50) - P(Z < - 0}}{\rm{.25)}}\\{\rm{ = 0}}{\rm{.6915 - 0}}{\rm{.4013}}\\{\rm{ = 0}}{\rm{.2902}}\\{\rm{ = 29}}{\rm{.02\% }}\end{array}\)

Therefore, the probability is \({\rm{P(12 < X < 15) = 0}}{\rm{.2902 = 29}}{\rm{.02\% }}\).

03

Calculating probability

b)

Given: \({\rm{X}}\) has a normal distribution with

\(\begin{array}{l}{\rm{\mu = 13\;min}}\\{\rm{\sigma = 4\;min}}\\{\rm{n = 16}}\end{array}\)

Since the distribution of \({\rm{x}}\)is normal, the sampling distribution of the sample mean\({\rm{\bar x}}\)is also normal.

The sampling distribution of the sample mean\({\rm{\bar x}}\)has mean \({\rm{\mu }}\) and standard deviation\(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\).

The z-score is the value decreased by the mean, divided by the standard deviation:

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{\bar x - }}{{\rm{\mu }}_{{\rm{\bar x}}}}}}{{{{\rm{\sigma }}_{{\rm{\bar x}}}}}}\\{\rm{ = }}\frac{{{\rm{12 - 13}}}}{{{\rm{4/}}\sqrt {{\rm{16}}} }}\\{\rm{\gg - 1}}{\rm{.00}}\\{\rm{z = }}\frac{{{\rm{\bar x - }}{{\rm{\mu }}_{{\rm{\bar x}}}}}}{{{{\rm{\sigma }}_{{\rm{\bar x}}}}}}\\{\rm{ = }}\frac{{{\rm{15 - 13}}}}{{{\rm{4/}}\sqrt {{\rm{16}}} }}\\{\rm{\gg 2}}{\rm{.00}}\end{array}\)

Determine the corresponding probability using the normal probability table in the appendix (which contain the probability to the left of z-scores:

\(\begin{array}{c}{\rm{P(12 < \bar X < 15) = P( - 1}}{\rm{.00 < Z < 2}}{\rm{.00)}}\\{\rm{ = P(Z < 2}}{\rm{.00) - P(Z < - 1}}{\rm{.00)}}\\{\rm{ = 0}}{\rm{.9772 - 0}}{\rm{.1587}}\\{\rm{ = 0}}{\rm{.8185}}\\{\rm{ = 81}}{\rm{.85\% }}\end{array}\)

Therefore, the probability is \({\rm{P(12 < \bar X < 15) = 0}}{\rm{.8185 = 81}}{\rm{.85\% }}\).

04

Calculating probability 

c)

Given:\({\rm{X}}\) has a normal distribution with

\(\begin{array}{l}{\rm{\mu = 13\;min}}\\{\rm{\sigma = 4\;min}}\end{array}\)

Result part (a):

\({\rm{P(12 < X < 15) = 0}}{\rm{.2902 = 29}}{\rm{.02\% }}\)

Result part (b):

\({\rm{P(12 < \bar X < 15) = 0}}{\rm{.8185 = 81}}{\rm{.85\% }}\)

Because the sample mean of multiple trucks is less variable than a single truck, the probability that the times are close to the mean of \({\rm{13}}\) minutes is much higher in part (b) than in part (a), and thus the probability that the times are close to the mean of \({\rm{13}}\) minutes is much higher in part (b) than in part (a).

05

Calculating probability

d)

Given: \({\rm{X}}\)has a normal distribution with

\(\begin{array}{l}{\rm{\mu = 13\;min}}\\{\rm{\sigma = 4\;min}}\\{\rm{n = 16}}\end{array}\)

Since the distribution of\({\rm{x}}\)is normal, the sampling distribution of the sample mean \({\rm{\bar x}}\) is also normal.

The sampling distribution of the sample mean \({\rm{\bar x}}\) has mean \({\rm{\mu }}\)and standard deviation\(\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}\).

The z-score is the value decreased by the mean, divided by the standard deviation:

\(\begin{array}{c}{\rm{z = }}\frac{{{\rm{\bar x - }}{{\rm{\mu }}_{{\rm{\bar x}}}}}}{{{{\rm{\sigma }}_{{\rm{\bar x}}}}}}\\{\rm{ = }}\frac{{{\rm{20 - 13}}}}{{{\rm{4/}}\sqrt {{\rm{16}}} }}\\{\rm{\gg 7}}{\rm{.00}}\end{array}\)

Determine the relevant probability using the appendix's normal probability table (which includes the probability to the left of z-scores:

\(\begin{array}{c}{\rm{P(\bar X > 20) = P(Z > 7}}{\rm{.00)}}\\{\rm{ = 1 - P(Z < 7}}{\rm{.00)}}\\{\rm{\gg 1 - 1 = 0}}\\{\rm{ = 0\% }}\end{array}\)

Therefore, the probability is \({\rm{P(\bar X > 20)\gg 0 = 0\% }}\).

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