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About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.

P ( x ≤ 14) = 0.9695

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In the 2013 Jerry’s Artarama art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X = the number of pages that feature signature artists.

  1. What values does x take on?
  2. What is the probability distribution? Find the following probabilities:
    1. the probability that two pages feature signature artists
    2. the probability that at most six pages feature signature artists
    3. the probability that more than three pages feature signature artists.
  3. Using the formulas, calculate the (i) mean and (ii) standard deviation.
  1. x = 0, 1, 2, 3, 4, 5, 6, 7, 8
  2. X ~ B ( 100 , 8 560 )
    1. P ( x = 2) = binompdf ( 100 , 8 560 , 2 ) = 0.2466
    2. P ( x ≤ 6) = binomcdf ( 100 , 8 560 , 6 ) = 0.9994
    3. P ( x >3) = 1 – P ( x ≤ 3) = 1 – binomcdf ( 100 , 8 560 , 3 ) = 1 – 0.9443 = 0.0557
    1. Mean = np = (100) ( 8 560 ) = 800 560 ≈ 1.4286
    2. Standard Deviation = n p q = ( 100 ) ( 8 560 ) ( 552 560 ) ≈ 1.1867
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According to a Gallup poll, 60% of American adults prefer saving over spending. Let X = the number of American adults out of a random sample of 50 who prefer saving to spending.

  1. What is the probability distribution for X ?
  2. Use your calculator to find the following probabilities:
    1. the probability that 25 adults in the sample prefer saving over spending
    2. the probability that at most 20 adults prefer saving
    3. the probability that more than 30 adults prefer saving
  3. Using the formulas, calculate the (i) mean and (ii) standard deviation of X .
  1. X B (50, 0.6)
  2. Using the TI-83, 83+, 84 calculator with instructions as provided in [link] :
    1. P ( x = 25) = binompdf(50, 0.6, 25) = 0.0405
    2. P ( x ≤ 20) = binomcdf(50, 0.6, 20) = 0.0034
    3. P ( x >30) = 1 - binomcdf(50, 0.6, 30) = 1 – 0.5535 = 0.4465
    1. Mean = np = 50(0.6) = 30
    2. Standard Deviation = n p q = 50 ( 0.6 ) ( 0.4 ) ≈ 3.4641
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The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let X = the number of people who will develop pancreatic cancer.

  1. What is the probability distribution for X ?
  2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X .
  3. Use your calculator to find the probability that at most eight people develop pancreatic cancer
  4. Is it more likely that five or six people will develop pancreatic cancer? Justify your answer numerically.
  1. X B (200, 0.0128)
    1. Mean = np = 200(0.0128) = 2.56
    2. Standard Deviation = n p q  =   (200)(0 .0128)(0.9872) 1. 5897
  2. Using the TI-83, 83+, 84 calculator with instructions as provided in [link] :
    P ( x ≤ 8) = binomcdf(200, 0.0128, 8) = 0.9988
  3. P ( x = 5) = binompdf(200, 0.0128, 5) = 0.0707
    P ( x = 6) = binompdf(200, 0.0128, 6) = 0.0298
    So P ( x = 5)> P ( x = 6); it is more likely that five people will develop cancer than six.
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During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let X = the number of shots that scored points.

  1. What is the probability distribution for X ?
  2. Using the formulas, calculate the (i) mean and (ii) standard deviation of X .
  3. Use your calculator to find the probability that DeAndre scored with 60 of these shots.
  4. Find the probability that DeAndre scored with more than 50 of these shots.
  1. X ~ B (80, 0.613)
    1. Mean = np = 80(0.613) = 49.04
    2. Standard Deviation = n p q = 80 ( 0.613 ) ( 0.387 ) 4.3564
  2. Using the TI-83, 83+, 84 calculator with instructions as provided in [link] :
    P ( x = 60) = binompdf(80, 0.613, 60) = 0.0036
  3. P ( x >50) = 1 – P ( x ≤ 50) = 1 – binomcdf(80, 0.613, 50) = 1 – 0.6282 = 0.3718
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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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