Respuesta :
Question 1:
Cheap-As-Dirt Rental Company advertises that the average cost of a rental they find for undergraduate students at the University of Oregon is $540 with a standard deviation of $75 (Let us assume the rents are approximately Normally distributed.) The Department of Consumer Protection will investigate the company if, when they choose a sample of students, they find that the average rental cost for those students is $610 or more.
Part (a):
Assuming that the company is advertising truthfully, what is the probability that the company will be investigated if the Department of Consumer Protection only samples one client?
The probability that an n sample of a normally distributed dataset with a mean, μ, and standard deviation, σ, exceeds a value x, is given by
[tex]P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \ \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \right)[/tex]
If one client is sampled, then
[tex]P(X\ \textgreater \ 610)=1-P\left(z\ \textless \ \frac{610-540}{\frac{75}{\sqrt{1}}} \right) \\ \\ =1-P\left(z\ \textless \ \frac{70}{\frac{75}{1}} \right)=1-P\left(z\ \textless \ \frac{70}{75} \right) \\ \\ =1-P(z\ \textless \ 0.9333)=1-0.82468=\bold{0.1753}[/tex]
Part (b):
Assuming that the company is advertising truthfully, the probability that the company will be investigated if the Department of Consumer Protection samples 10 clients is given by:
[tex]P(X\ \textgreater \ 610)=1-P\left(z\ \textless \ \frac{610-540}{\frac{75}{\sqrt{10}}} \right) \\ \\ =1-P\left(z\ \textless \ \frac{70}{\frac{75}{3.162}} \right)=1-P\left(z\ \textless \ \frac{70}{23.72} \right) \\ \\ =1-P(z\ \textless \ 2.951)=1-0.99842=\bold{0.0016}[/tex]
Part (c):
Assuming that the company is advertising truthfully, the probability that the company will be investigated if the Department of Consumer Protection samples 50 clients is given by:
[tex]P(X\ \textgreater \ 610)=1-P\left(z\ \textless \ \frac{610-540}{\frac{75}{\sqrt{50}}} \right) \\ \\ =1-P\left(z\ \textless \ \frac{70}{\frac{75}{7.071}} \right)=1-P\left(z\ \textless \ \frac{70}{10.61} \right) \\ \\ =1-P(z\ \textless \ 6.5997)=1-1=\bold{0}[/tex]
Question 2. In New York City in the late 1970s and early 1980s, parking meter collections averaged $1.75 million per month with a standard deviation of $302 thousand per month. Assume that the amount generated per month is Normal with the mean and standard deviation from the previous sentence. For about two years the city hired Brink’s Inc. to do collections instead. The data file gives the collections that Brink’s reported to the city during that period.
Part (a):
The sample size for this data is the number of months in two years which is 24 months.
Part (b):
The sample mean is given by the average of the data in the datafile. This is obtained by adding the data in the datasheet and dividing the result by 24.
Thus sample mean is given by
[tex](1330143+1525370+1312257+1494717+1385822+1557598+ \\ 1469066+1547011+1565671+1124576+1773806+1652490+ \\ 1744558+1678286+1636597+1738847+1576383+1846508+ \\ 1746899+1685611+1794467+1695017+1671626+1566107) / 24 \\ \\ = \frac{38,119,433}{24} =\$1,588,309.71[/tex]
Part (c):
The city was concerned that Brink’s was stealing money from the parking meter receipts. Assuming that this sample of months is like a random sample. Given the answers to the first two parts, the probability of a sample at least as far below the mean as the sample from the period of the Brink’s contract is given by
[tex]P(X\ \textless \ 1,588,309.71)=P\left(z\ \textless \ \frac{1,588,309.71-1,750,000}{\frac{302,000}{\sqrt{24}}} \right) \\ \\ =P\left(z\ \textless \ \frac{-161,690.29}{\frac{302,000}{4.899}} \right)=P\left(z\ \textless \ \frac{-161,690.29}{61,645.49} \right)=P(z\ \textless \ -2.623) \\ \\ =\bold{0.0044}[/tex]
Cheap-As-Dirt Rental Company advertises that the average cost of a rental they find for undergraduate students at the University of Oregon is $540 with a standard deviation of $75 (Let us assume the rents are approximately Normally distributed.) The Department of Consumer Protection will investigate the company if, when they choose a sample of students, they find that the average rental cost for those students is $610 or more.
Part (a):
Assuming that the company is advertising truthfully, what is the probability that the company will be investigated if the Department of Consumer Protection only samples one client?
The probability that an n sample of a normally distributed dataset with a mean, μ, and standard deviation, σ, exceeds a value x, is given by
[tex]P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \ \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \right)[/tex]
If one client is sampled, then
[tex]P(X\ \textgreater \ 610)=1-P\left(z\ \textless \ \frac{610-540}{\frac{75}{\sqrt{1}}} \right) \\ \\ =1-P\left(z\ \textless \ \frac{70}{\frac{75}{1}} \right)=1-P\left(z\ \textless \ \frac{70}{75} \right) \\ \\ =1-P(z\ \textless \ 0.9333)=1-0.82468=\bold{0.1753}[/tex]
Part (b):
Assuming that the company is advertising truthfully, the probability that the company will be investigated if the Department of Consumer Protection samples 10 clients is given by:
[tex]P(X\ \textgreater \ 610)=1-P\left(z\ \textless \ \frac{610-540}{\frac{75}{\sqrt{10}}} \right) \\ \\ =1-P\left(z\ \textless \ \frac{70}{\frac{75}{3.162}} \right)=1-P\left(z\ \textless \ \frac{70}{23.72} \right) \\ \\ =1-P(z\ \textless \ 2.951)=1-0.99842=\bold{0.0016}[/tex]
Part (c):
Assuming that the company is advertising truthfully, the probability that the company will be investigated if the Department of Consumer Protection samples 50 clients is given by:
[tex]P(X\ \textgreater \ 610)=1-P\left(z\ \textless \ \frac{610-540}{\frac{75}{\sqrt{50}}} \right) \\ \\ =1-P\left(z\ \textless \ \frac{70}{\frac{75}{7.071}} \right)=1-P\left(z\ \textless \ \frac{70}{10.61} \right) \\ \\ =1-P(z\ \textless \ 6.5997)=1-1=\bold{0}[/tex]
Question 2. In New York City in the late 1970s and early 1980s, parking meter collections averaged $1.75 million per month with a standard deviation of $302 thousand per month. Assume that the amount generated per month is Normal with the mean and standard deviation from the previous sentence. For about two years the city hired Brink’s Inc. to do collections instead. The data file gives the collections that Brink’s reported to the city during that period.
Part (a):
The sample size for this data is the number of months in two years which is 24 months.
Part (b):
The sample mean is given by the average of the data in the datafile. This is obtained by adding the data in the datasheet and dividing the result by 24.
Thus sample mean is given by
[tex](1330143+1525370+1312257+1494717+1385822+1557598+ \\ 1469066+1547011+1565671+1124576+1773806+1652490+ \\ 1744558+1678286+1636597+1738847+1576383+1846508+ \\ 1746899+1685611+1794467+1695017+1671626+1566107) / 24 \\ \\ = \frac{38,119,433}{24} =\$1,588,309.71[/tex]
Part (c):
The city was concerned that Brink’s was stealing money from the parking meter receipts. Assuming that this sample of months is like a random sample. Given the answers to the first two parts, the probability of a sample at least as far below the mean as the sample from the period of the Brink’s contract is given by
[tex]P(X\ \textless \ 1,588,309.71)=P\left(z\ \textless \ \frac{1,588,309.71-1,750,000}{\frac{302,000}{\sqrt{24}}} \right) \\ \\ =P\left(z\ \textless \ \frac{-161,690.29}{\frac{302,000}{4.899}} \right)=P\left(z\ \textless \ \frac{-161,690.29}{61,645.49} \right)=P(z\ \textless \ -2.623) \\ \\ =\bold{0.0044}[/tex]
