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The next three questions refer to the following information:

In a survey at Kirkwood Ski Resort the following information was recorded:

Sport participation by age
0 – 10 11 - 20 21 - 40 40+
Ski 10 12 30 8
Snowboard 6 17 12 5

Suppose that one person from of the above was randomly selected.

Find the probability that the person was a skier or was age 11 – 20.

77 100 size 12{ { { size 8{"77"} } over { size 8{"100"} } } } {}

Find the probability that the person was a snowboarder given he/she was age 21 – 40.

12 42 size 12{ { { size 8{"12"} } over { size 8{"42"} } } } {}

Explain which of the following are true and which are false.

  • Sport and Age are independent events.
  • Ski and age 11 – 20 are mutually exclusive events.
  • P ( Ski and age 21 40 ) < P ( Ski age 21 40 ) size 12{P \( ital "Ski"+ ital "age""21" - "40" \)<P \( ital "Ski" \lline ital "age""21" - "40" \) } {}
  • P ( Snowboard or age 0 10 ) < P ( Snowboard age 0 10 ) size 12{P \( ital "Snowboardorage"0 - "10" \)<P \( ital "Snowboard" \lline ital "age"0 - "10" \) } {}
  • False
  • False
  • True
  • False

The average length of time a person with a broken leg wears a cast is approximately 6 weeks. The standard deviation is about 3 weeks. Thirty people who had recently healed from broken legs were interviewed. State the distribution that most accurately reflects total time to heal for the thirty people.

N ( 180 , 16 . 43 ) size 12{N \( "180","16" "." "43" \) } {}

The distribution for X size 12{X} {} is Uniform. What can we say for certain about the distribution for X ¯ size 12{ {overline {X}} } {} when n = 1 size 12{n=1} {} ?

  • The distribution for X ¯ size 12{ {overline {X}} } {} is still Uniform with the same mean and standard dev. as the distribution for X size 12{X} {} .
  • The distribution for X ¯ size 12{ {overline {X}} } {} is Normal with the different mean and a different standard deviation as the distribution for X size 12{X} {} .
  • The distribution for X ¯ size 12{ {overline {X}} } {} is Normal with the same mean but a larger standard deviation than the distribution for X size 12{X} {} .
  • The distribution for X ¯ size 12{ {overline {X}} } {} is Normal with the same mean but a smaller standard deviation than the distribution for X size 12{X} {} .

A

The distribution for X size 12{X} {} is uniform. What can we say for certain about the distribution for X size 12{ Sum {X} } {} when n = 50 size 12{n=50} {} ?

  • The distribution for X size 12{ Sum {X} } {} is still uniform with the same mean and standard deviation as the distribution for X size 12{X} {} .
  • The distribution for X size 12{ Sum {X} } {} is Normal with the same mean but a larger standard deviation as the distribution for X size 12{X} {} .
  • The distribution for X size 12{ Sum {X} } {} is Normal with a larger mean and a larger standard deviation than the distribution for X size 12{X} {} .
  • The distribution for X size 12{ Sum {X} } {} is Normal with the same mean but a smaller standard deviation than the distribution for X size 12{X} {} .

C

The next three questions refer to the following information:

A group of students measured the lengths of all the carrots in a five-pound bag of baby carrots. They calculated the average length of baby carrots to be 2.0 inches with a standard deviation of 0.25 inches. Suppose we randomly survey 16 five-pound bags of baby carrots.

State the approximate distribution for X ¯ size 12{ {overline {X}} } {} , the distribution for the average lengths of baby carrots in 16 five-pound bags. X ¯ ~ size 12{ {overline {X}} "~" } {}

N ( 2 , .25 16 ) size 12{N \( { { size 8{2 "." "25"} } over { size 8{ sqrt {"16"} } } } \) } {}

Explain why we cannot find the probability that one individual randomly chosen carrot is greater than 2.25 inches.

Find the probability that x ¯ size 12{ {overline {x}} } {} is between 2 and 2.25 inches.

0.5000

The next three questions refer to the following information:

At the beginning of the term, the amount of time a student waits in line at the campus store is normally distributed with a mean of 5 minutes and a standard deviation of 2 minutes.

Find the 90th percentile of waiting time in minutes.

7.6

Find the median waiting time for one student.

5

Find the probability that the average waiting time for 40 students is at least 4.5 minutes.

0.9431

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Source:  OpenStax, Elementary statistics. OpenStax CNX. Dec 30, 2013 Download for free at http://cnx.org/content/col10966/1.4
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