ENGI 3423 Probability and Statistics

Faculty of Engineering and Applied Science
2008 Fall

Final Examination Questions

  1. Events A and B are such that   P[A] = .3,   P[B] = .2   and the probability that exactly one of the two events occurs is .38 .

    1. Find the probability that both events A and B occur.

      [6]

    2. Are the events   A, B
      1. incompatible (mutually exclusive) but not independent?
      2. independent but not incompatible?
      3. both independent and incompatible?   or
      4. neither independent nor incompatible?
      [Circle whichever one of (i), (ii), (iii) or (iv) is correct and justify your choice.]

      [4]

  1. A team for an engineering competition consists of two Term 3 students and three Term 6 students.   The five team members have equal status.   Twenty students from Term 3 and fifteen students from Term 6 have volunteered to serve on the team.   In how many distinct ways can the team be chosen from among the volunteers?   [Your final answer must be a single whole number.]

    [10]

  1. From prior experience, it is believed that the mean load at failure of a certain type of beam is 10 000 N.   The strength of this belief is represented by the standard deviation sigma0 = 2 N.   A random sample of 64 such beams produces a sample mean failure load of 9995.0 N with a sample standard deviation of 16.0 N.

    1. Construct a Bayesian 95% confidence interval estimate for the true mean failure load   µ.

      [8]

    2. Construct a classical 95% confidence interval estimate for the true mean failure load   µ.

      [5]

    3. Is there sufficient evidence to conclude that the true mean failure load   µ   is not 10 000 N?

      [2]

  1. The lifetimes of two types of lamps are known to be normally distributed random quantities, with mean muA = 950 hours and standard deviation sigmaA = 40 hours for type A and mean muB = 1000 hours and standard deviation sigmaB = 30 hours for type B.   One lamp of each type is chosen at random.

    1. Find the probability that the lamp of type B lasts at least 100 hours longer than the lamp of type A.

      [9]

    2. Find the probability that the sum of the two lifetimes exceeds 2000 hours.

      [6]

  1. The time (in minutes) needed for a random sample of 21 machines to complete a certain task is measured both before and after a refit.   A consultant claims that the refit decreases the time needed to complete the task.   It is known that both populations are normally distributed, with a common variance.

    1. Which of the two sample t-tests (paired or unpaired) should be conducted?
      State the reason for your selection.

      [3]

    2. Conduct the appropriate hypothesis test, at a level of significance of 1%, using whichever one of the two sets of Minitab® output is appropriate:

      Two-sample T for Before vs After

         N   Mean  StDev  SE Mean
Before  21  62.46   3.83     0.84
After   21  60.22   4.11     0.90

Difference = mu (Before) - mu (After)
Estimate for difference:  2.25
99% lower bound for difference:  -0.72
T-Test of difference = 0 (vs >): T-Value = 1.83  P-Value = 0.037  DF = 40
Both use Pooled StDev = 3.9722

      

Paired T for Before - After

             N    Mean  StDev  SE Mean
Before      21  62.463  3.827    0.835
After       21  60.217  4.112    0.897
Difference  21   2.246  3.555    0.776

99% lower bound for mean difference: 0.285
T-Test of mean difference = 0 (vs > 0): T-Value = 2.90  P-Value = 0.004
Both use Pooled StDev = 23.4053

      [10]

    3. Does the evidence support the consultant’s claim?

      [2]

  1. The deceleration (y mm/s2) of test spheres of radius (x mm) falling through a viscous medium is measured as they reach a certain speed.   The following summary statistics are known.
    Sum x = 2431, Sum x^2 = 269 511, Sum xy = 27 019.564 n = 22, Sum y = 243.663, Sum y^2 = 2 712.477
    These values lead to
    n Sxx = 19 481, n Sxy = 2 085.655, n Syy = 302.836

    1. Show that the equation of the simple linear regression line for these data (y on x) is
              y   =   0.107 1 x – 0.754 6   (correct to 4 s.f. in each coefficient).

      [4]

    2. Complete an ANOVA table for these data.

      [5]

    3. To the nearest 1%, how much of the total variation in   Y   is explained by the simple linear regression model?

      [3]

    4. Conduct an appropriate hypothesis test to determine whether or not there is a significant linear association between   Y   and   x.

      [3]

    5. Construct a 95% prediction interval for a future observation of   Y   when   x = 110.5 .

      [5]

  1. The random quantity   X   has the probability density function

    f(x) = k sech^2 x

    1. Starting from the p.d.f.   f (x) , show that the cumulative distribution function for   X   is   F(x) = (1 + tanh x) / 2   and hence find the value of   k .

      [6]

    2. Find the value of the median   median (mu~) .

      [3]

    3. Show that the upper quartile is at   xU = (ln 3) / 2 .

      [6]

    [Note:   You may quote   tanh x = (e^x - e^(-x)) / (e^x + e^(-x)) and d/dx(tanh x) = sech^2 x .
    Also note that   sech x = 1 / cosh x = 2 / (e^x + e^(-x)) .]

  1. BONUS QUESTION:

    A plastic sheath is being constructed to enclose a new telecommunications cable.   Before a section of the cable is accepted for use, the sheath may be inspected for the presence of defects.   The result is either a pass (P) or failure to pass the test (F = ~P).   From past experience, it is estimated that 80% of all sections constructed by the contractor are free of defects.   There are two states of nature to be considered:   good (defect-free, G), or defect present (D = ~G).   The test is not perfect; there is a 5% chance that the test gives the result P when there is a defect present and there is a 10% chance that the test gives the result F when there is no defect.

    A particular section of the cable is under investigation for possible defects.   Consequences are measured on a monetary scale ranging from $1000 to –$4130, as indicated below.   A decision analysis has been decided upon to assist decision-making.   Construct a decision tree, showing two possible strategies, as follows.

    Action 1:   {do not test; accept the cable section}.
    The consequences in this case are $1000 if the section is good,
              –$4000 if defective.
    Action 2:   {test; if the result is a pass then accept without remedial action,
              otherwise take remedial action and accept}.
    The cost of testing is $130.
    The cost of taking remedial action is $500.
    The consequences in this case are:
              $1000 if the section is good,
              –$4000 if the section is defective and no remedial action is taken, and
              –$500 if the section is defective but remedial action is taken.
    These values should be adjusted to reflect the cost of testing and the cost of taking remedial action where appropriate.   Determine which strategy is the optimal strategy and find the expected gain from the optimal strategy.

    [+10]

[Also provided with this examination paper were tables of the standard normal c.d.f. (the z tables)
    and of the critical values of the t distribution (the t tables).]

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Created 2008 12 16 and most recently modified 2008 12 18 by Dr. G.H. George.