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Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of 7 days. We randomly sample 9 trials.

  • In words, ΣX = size 12{ΣX={}} {}
  • ΣX ~ size 12{ΣX "~" } {}
  • Find the probability that the total length of the 9 trials is at least 225 days.
  • 90 percent of the total of 9 of these types of trials will last at least how long?
  • The total length of time for 9 criminal trials
  • N ( 189 , 21 ) size 12{ ital "SumX" "~" N \( 9 * "21", sqrt {9} * 7 \) } {}
  • 0.0432
  • 162.09
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According to the Internal Revenue Service, the average length of time for an individual to complete (record keep, learn, prepare, copy, assemble and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is 2 hours. Suppose we randomly sample 36 taxpayers.

  • In words, X = size 12{X={}} {}
  • In words, X ¯ = size 12{ {overline {X}} ={}} {}
  • X ¯ ~ size 12{ {overline {X}} "~" } {}
  • Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.
  • Would you be surprised if one taxpayer finished his Form 1040 in more than 12 hours? In a complete sentence, explain why.
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Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races.

Let X ¯ = size 12{ {overline {X}} ={}} {} the average of the 49 races.

  • X ¯ ~ size 12{ {overline {X}} "~" } {}
  • Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.
  • Find the 80th percentile for the average of these 49 marathons.
  • Find the median of the average running times.
  • N ( 145 , 14 49 ) size 12{ ital "Xbar" "~" N \( "145", { {"14"} over { sqrt {"49"} } } \) } {}
  • 0.6247
  • 146.68
  • 145 minutes
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The attention span of a two year-old is exponentially distributed with a mean of about 8 minutes. Suppose we randomly survey 60 two year-olds.

  • In words, X = size 12{X={}} {}
  • X ~ size 12{X "~" } {}
  • In words, X ¯ = size 12{ {overline {X}} ={}} {}
  • X ¯ ~ size 12{ {overline {X}} "~" } {}
  • Before doing any calculations, which do you think will be higher? Explain why.
    • the probability that an individual attention span is less than 10 minutes; or
    • the probability that the average attention span for the 60 children is less than 10 minutes? Why?
  • Calculate the probabilities in part (e).
  • Explain why the distribution for X ¯ size 12{ {overline {X}} } {} is not exponential.
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Suppose that the length of research papers is uniformly distributed from 10 to 25 pages. We survey a class in which 55 research papers were turned in to a professor. The 55 research papers are considered a random collection of all papers. We are interested in the average length of the research papers.

  • In words, X = size 12{X={}} {}
  • X ~ size 12{X "~" } {}
  • μ X = size 12{μ rSub { size 8{X} } ={}} {}
  • σ X = size 12{σ rSub { size 8{X} } ={}} {}
  • In words, X ¯ = size 12{ {overline {X}} ={}} {}
  • X ¯ ~ size 12{ {overline {X}} "~" } {}
  • In words, ΣX = size 12{ΣX={}} {}
  • ΣX ~ size 12{ΣX "~" } {}
  • Without doing any calculations, do you think that it’s likely that the professor will need to read a total of more than 1050 pages? Why?
  • Calculate the probability that the professor will need to read a total of more than 1050 pages.
  • Why is it so unlikely that the average length of the papers will be less than 12 pages?
  • U ( 10 , 25 ) size 12{X "~" U \( "10","25" \) } {}
  • 17.5
  • 225 12 = 4.3301
  • N ( 17.5 , 0.5839 ) size 12{X "~" U \( "10","25" \) } {}
  • N ( 962.5 , 32.11 ) size 12{X "~" U \( "10","25" \) } {}
  • 0.0032
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The length of songs in a collector’s CD collection is uniformly distributed from 2 to 3.5 minutes. Suppose we randomly pick 5 CDs from the collection. There is a total of 43 songs on the 5 CDs.

  • In words, X = size 12{X={}} {}
  • X ~ size 12{X "~" } {}
  • In words, X ¯ = size 12{ {overline {X}} "~" } {}
  • X ¯ ~ size 12{ {overline {X}} "~" } {}
  • Find the first quartile for the average song length.
  • The IQR (interquartile range) for the average song length is from _______ to _______.
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Questions & Answers

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4
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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Source:  OpenStax, Collaborative statistics. OpenStax CNX. Jul 03, 2012 Download for free at http://cnx.org/content/col10522/1.40
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