MAT 470, Worksheet 3 Solutions

  1. Given the density function…
    a). Determine the probability, P( Y < 0.5 ). b). Determine the probability, P( Y < 2 ).
    This region is triangular with area Sum the triangular and trapezoid areas
    d). Determine the cumulative distribution function, F(Y).
  2. Entering a “T-intersection”, a driver must choose to turn either left or right. It’s a single lane road, so only one car at a time may enter the intersection from this direction. The probability any of these cars makes a left turn is 0.40.
    a). How many cars do you expect to observe entering the intersection to find the first car
    that makes a left turn?
    For the geometric, mean equals 1/p = 2.5
    b). If you observe 6 cars pass through the intersection, what is the probability exactly 2 of
    these cars turn left?
    P( Y = 6) = binompdf(6, 0.4, 2) = 0.31104
    c). If you observe 6 cars pass through the intersection, what is the probability at least 2 of
    these cars turn left?
    Using binomial, P(Y 2) = 1 – P(Y 1) = 1 – P( Y = 0) – P( Y = 1 ) = 0.76672
    d). At this intersection, the traffic signal stays red for one minute. On average, 5 cars arrive
    at the intersection during the red light. What is the probability only 3 cars arrive during
    a red light?
    P( X = 3 ) = poissonpdf(5, 3) = 0.14037
    e). If you just observed 3 cars in a row make a right turn at the intersection, what is the
    probability you need to observe at least 4 more cars to find the first one that turns left?
    The geometric is memoryless and so P( Y 7 | Y > 3) = P( Y 4 ) = q3 = 0.216
  3. Of the volunteers donating blood in a clinic, 80% have the Rhesus factor present in their blood.
    a). If 8 of the volunteers are randomly selected, what is the probability that exactly 6
    have the Rhesus factor?
    P( Y = 6 ) = binompdf(8, 0.80, 6 ) = 0.2936
    b). If 8 of the volunteers are randomly selected, what is the probability that at least 6
    have the Rhesus factor?
    compute and sum P(Y = 6 ) + P( Y = 7 ) + P( Y = 8 ) = 0.7969
    c). If volunteers are screened and tested one at a time, what is the probability you test 5
    individuals before finding the first one without the Rhesus factor? (now, let p = 0.2)
    For the geometric, P(Y = 5) = q4p = (.8)4(.2) = 0.08192
    d). For samples of 8 volunteers, what is the mean number of individuals without the
    Rhesus factor?
    For the binomial with 8 trials, the mean is np = (8)(.2) = 1.6
  4. A oil company is preparing to drill exploratory holes in search of oil. Suppose the probability of finding oil on each individual exploration is 0.15 and the explorations are independent.
    a). If a total of 12 holes are drilled, what is the probability that exactly 3 are successful?
    The binomial probability is given by
    P ( Y = 3 ) = C12, 3 (0.15)3 (0.85)9 = 0.17198
    b). If a total of 12 holes are drilled, what is the probability that more than 3 are successful?
    P(Y > 3) = 1 – P(Y 3) = 1 –P(Y = 0) – P(Y = 1) – P(Y = 2) – P(Y = 3) = 0.092206
    c). If only one exploration is done at a time, what is the probability the first successful
    drilling occurs on the third exploration?
    Using the geometric distribution, P(Y = 3) = (0.85)2 (0.15) = 0.108375
    d). If only one exploration is done at a time, what is the probability more than six holes
    are drilled before the first success exploration occurs?
    Using the geometric distribution,
    P(Y > 6) = P( first six are failures) = (0.85)6 = 0.37715
    e). Suppose the company continues to drill holes until a success occurs.
    What is the mean number of holes required to achieve a success?
    For a geometric distribution, the mean is E(Y) = 1/p = 1 / 0.15 = 100/15
  5. Suppose for each highway mile you drive, there is a probability of 0.05 that you see a dead deer along the roadside. Assume these occurrences are independent. Consider driving 100 miles of highway. Based on the binomial distribution, what is the probability of observing exactly 8 dead deer during your 100-mile drive?
    P( Y = 8 ) = binomialpdf( 100, 0.05, 8 ) = 0.06487
  6. Car arrivals at two entrances to a parking lot are independent. Arrivals at Entrance A have a Poisson distribution with an average of 3 per hour. Arrivals at Entrance B have a Poisson distribution with an average of 5 per hour. In a given one-hour period, find the probability that 4 cars arrive by Entrance A and 6 cars arrive by Entrance B.
    Since independent events, we know P( A = 4 and B = 6 ) = P(A = 4)P(B = 6).
    Therefore,
  7. In an assembly line, gearbox assemblies can be installed in one minute each if the holes are properly drilled, but the installation takes ten minutes if the holes need to be redrilled. Thirty gearboxes are in stock, 6 with improperly drilled holes. Five of the 30 gearboxes are selected to be used in the next 5 installations.
    a). Find the probability exactly 4 of the selected gearboxes will fit properly.
    Use the hypergeometric distribution with N = 30, r = 24, and n = 5.
    b). Find the expected value for the number of gearboxes that fit properly when 5 are
    selected.
    E(Y) = (nr)/N = 4
    c). Find the expected value for the time it takes to install these 5 gearboxes.
    E(T) = 4 ( 1 minute ) + 1 (10 minutes ) = 14 minutes.
  8. Suppose among a supply of 5000 parts produced during a given week, there are 100 that don’t meet the required quality standard. Twenty of the parts are randomly selected and checked to see if they meet the standard. Let Y be the number in the sample that don’t meet the standard.
    a). Compute the probability exactly 2 of the sampled parts fail to meet the quality standard.
    Here, N = 5000, n = 20, r = 100, and so
    b). Determine the mean, E(Y), and the variance, V(Y) .
    E(Y) = 20(100)/5000 = 2/5 = 0.4