STA 6166 UNIT 1 Section 3 Answers
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Unit 1 Section 3 Answers

You can choose to work some or all of the problems listed below. We recommend that you at least work the general questions and the problems listed in your major area of interest. When you finish, check yourself against the answers. Answers can be found here.

General Questions.
  1. Identify the following random variables as being either discrete or continuous.
    1. Concentration of an environmental toxin in eggshells of a given bird species.[Continuous]
    2. Number of mutations in a bacterial culture after exposure to a mutagen.[Discrete]
    3. Understory biomass in a quarter meter survey area.[Continuous]
    4. Blood pressure of a heart patient in a clinical trial.[Continuous]
    5. Age of the same heart patient.[Discrete]
    6. Time to failure in a concrete pressure test.[Continuous]
    7. Proportion of citrus fruit on a tree that is infected by the Mediterranean Fruit fly larvae.[Continuous]
    8. Number of high school graduates in a sample of 100 prisoners.[Discrete]
  2. For each of the following pairs of events, decide whether or not the events are independent. Can you explain why?
    1. The number of eggs laid (A) and the concentration of toxins in the eggshell (B) for birds living in a contaminated ecosystem.[Not independent - Higher concentrations in the eggshell would be indicator of hight concentrations in the mother. Higher contaminated mother less likely to have large numbers of eggs.]
    2. Failure of a metal beam (A) and the size of the load imposed on it (B).[Not independent - the higher the load the more likely the beam is to fail, assuming all beams are of the same size.]
    3. The height (A) and weight (B) of an individual. [Would be expected to not be independent since the taller the individual the more weight they are expected to have. Note that the relationship does not have to be exact for independence to be rejected.]
    4. The age (A) and gender (B) of an individual. [Would be expected to be independent except at the extreme old age - where females tend to live longer. At most other ages, there is a 50:50 split in the population.]
    5. The number of telephone poles in a county (A) and the number of lung cancer deaths per year (B). [We would expect these to be independent since one would have to work very hard to come up with a hypothesis to explain a relationship between the two. Nevertheless, there is a significant correlation between these two variables, hence statistically they are not independent.]
  3. Work problem 4.28 in Ott and Longnecker, page 140. Can you rewrite this problem for a common situation in your area of study?
  4. Which of the following are valid probability (mass) density functions?
    1. x
      0
      2
      4
      6
      8
      P(X=x)
      0.2
      0.2
      0.2
      0.2
      0.2
      Yes this is a pdf.
    2. x
      -2
      -1
      0
      1
      P(X=x)
      0.22
      0.31
      0.23
      0.24
      Yes this is a pdf.
    3. X
      0
      1
      2
      3
      4
      5
      6+
      P(X=x)
      0.06
      0.21
      0.33
      0.31
      0.21
      -0.19
      0.08
      No way! You can never have a probability less than zero.
  5. Let Y be a binomial random variable. Compute the following probabilities. [ All can be solved with the equation from page 146. Note that for part c here, you have to add up the probabilities from zero to 12 (13 not included).]
    1. Let n=10, p=0.2, compute P(Y=3).[0.201]
    2. Let n=4, p=0.4, compute P(Y=2).[0.346]
    3. Let n=16, p=0.7, compute P(Y<13).[0.754]
  6. Use Table 1 in the Appendix of Ott and Longnecker to find the area under the standard normal curve for the following.
    1. For z between 0.0 and z=1.3 [0.9032-0.5 = 0.4032]
    2. For z between -1.0 and z=1.0 [0.8413 - 0.1587 = 0.6826]
    3. For z between 0.0 and z= -1.3 [.5000 - 0.0968 = 0.4032]
    4. For z greater than 1.75 [1.0 - .9599 = 0.0401 ]
    5. For z less than -1.75 [ Same as d by symmetry ]
  7. Using Table 1, find the value of z, call it z0, such that P(Z>z0) = 0.25. [ .677 ]
  8. Suppose that Y is a normal random variable with m = 100 and s = 15. Compute the following using Table 1 in the Appendix.
    1. P(Y<100) [ = P(z<(100-m)/s)=P(z<0) = 0.5000 ]
    2. P(Y>110) [ = P(z>(110-100)/15) = P(z>0.67) = 1.0-0.7486=0.2514 ]
    3. P(88 < Y < 120) [ = P((88-100)/15 <z < (120-100)/15) = P( -0.8<z<1.33) = P(z < 1.33)-P(z<-.8) = 0.9082-0.2119 = 0.6963 ]
    4. P(100 < Y < 108) [=P(z<(108-100)/15)-P(z<(100-100)/15) = P(z<0.53)-P(z<0) = 0.7019-0.5000 = 0.2019 ]
    5. Find the value k such that P(100 - k < Y < 100 + k) = 0.6 [ k=12.6 check it out for yourself. ]
  9. State in your own words what the Central Limit Theorem for the Sample Mean says. [I would never put words in your mouth.]
For students in agriculture and environmental fields.
  1. As part of a wading bird research project in the Florida Everglades, you monitor nesting in two large egret rookeries for egg laying and nesting success. Results are described in the table below.
    2.  
    Number of Nests Examined
    Number of Nests with Eggs
    Number of Nests with Hatchlings
    Rookery 1
    103
    37
    17
    Rookery 2
    92
    21
    9
    Total
    195
    58
    26

    Using this table compute the following:

    1. What is the (estimate of the) probability of finding an egg in a nest? [58/195 = 0.297 ]
    2. What is the conditional probability of finding an egg in a nest given it is from Rookery 1? [37/103 = 0.359 ]
    3. What is the probability of a nest producing a hatchling? [ 26/195 = 0.133 ]
    4. What is the conditional probability of a nest producing a hatchling given that the nest has been reported as having at least one egg? [ 26/58 = 0.448 ]
    5. Are the estimates of the conditional probability of a nest producing a hatchling the same for Rookery 1 as for Rookery 2. [ Rookery 2 conditional probability is 21/92 = 0.228 which is less than the 0.359 of Rookery 1. It would be nice to have a statistical test here to determine the probability that these numbers are really different.]

     

  2. Yield of a particular fruit is known to be normally distributed with mean of 10 kg/tree and standard deviation of 2 kg/tree. Suppose yields of twenty trees are to be collected and the average computed. What is the probability of observing an average yield below 9 kg/tree? (HINT: use the Central Limit Theorem to tell you what the sampling distribution of the mean should be, compute the z-score for 9 kg/tree, then find the appropriate probability using Table 1).

    3. The problem can be formulated as follows: Let Y be the random variable representing average tree yield. We want to compute P(Y < 9). Now, if Y is an average from a population with mean 10 and standard deviation 2, the Central Limit Theorem says that the new standardized random variable Z = (Y-10)/(2/sqrt(n)) has a standard normal distribution. The n here is 20. Hence we want to know P(Z < (9-10)/(2/sgrt(20))) or P(Z < -2.23) = 0.0129.

For students in engineering fields.
  1. The emergency room of a hospital has two large backup generators, either of which can supply sufficient electricity for basic operations in the event of loss of power from the regional grid. Each generator is tested a number of times over a year with the results given below;
    2. Number of times tested
    Number of times it failed to start
    Generator 1
    104
    3
    Generator 2
    104
    6
    Total
    208
    9

    With this information compute the following:

    1. What is the (estimate of the ) probability that Generator 1 will fail to start? [ 3/104 = 0.029 ]
    2. What is the probability that Generator 1 will start? [(104-3)/104 = 0.971 ]
    3. What is the probability that one or the other of the generators will work when needed ( P(Gen 1 works or Gen 2 works)? [ This is not as simple a problem as it initially seems. First, we need to make an assumption, mainly that the successful start of Gen 1 is independent of the successful start of Gen 2. Now, let A be the event that Gen 1 starts and let B be the event that Gen 2 starts. P(A or B) = P(A) + P(B) - P(A and B). P(A) = 101/104 = .9711. P(B) = 98/104 = 0.9423. P(A and B) = P(A)P(B) =0.9151 (by independence). Hence P(A or B) = .9711 + .9423 - .9151 = 0.9983 ]
    4. What is the probability that both generators fail to work simultaneously when needed ( P(Gen 1 fails and Gen 2 fails) [ If we can assume that A is independent of B, then it follows that the complement of A should be independed of the complement of B. Let C be the event that Gen 1 fails (complement of A) and let D be the event that Gen 2 fails (complement of B). Then P(C and D) = P(C)P(D) = (3/104)(6/104)=0.0017 = 1-0.9983 that is the probability that both fail simultaneously is 1 minus the probability that at least one of then successfully starts.]
  2. A robotic device for tightening a bolt is designed to produce torque values that are normally distributed with mean of 8 ft-lbs and standard deviation of 1 ft-lbs. We plan to tighten 30 bolts with this device, then measure the actual torque level. What is the probability of observing an average torque value for these 30 readings that is between 7.4 ft-lbs and 8.6 ft-lbs. (HINT: use the Central Limit Theorem to tell you what the sampling distribution of the mean should be, compute the z-scores for the torque limits, then find the appropriate probability using Table 1).

    3. Let us formulate the problem as follows. Let T be a random variable representing the AVERAGE torque value for these 30 readings. Then T is a mean from 30 samples from a population with expected mean of 8 and standard deviation of 1. By the Central Limit Theorem, Z = (T-8)/(1/sqrt(30)) will have a standard normal distribution. We want to know then
For students in toxicology and health science fields.
  1. A simple bioassay is performed to determine the toxicity of a pesticide on a infaunal copepod. The marker of aquatic toxicity was the capacity of the copepod to produce young that grow to the adult stage (26-day maturation). The table below presents information on the numbers of survivors to adulthood for the control group (no pesticide) and the treated group (40 microgram/l exposure).
    2.  
    Sex of Offspring
     
    Female
    Male
    Total
    Control
    315
    107
    422
    Treatment
    229
    93
    322
    Total
    544
    200
    744

    Compute the following:

    1. What is the probability of observing a male offspring (P(event A=being male)? [200/744 = 0.269 ]
    2. What is the conditional probability of observing a male offspring given the control group (P(A|B=being in the control group))? [ 107/422 = 0.2535 ]
    3. What is the probability of a male offspring being also part of the control treatment group (P(A and B)? [ 107/744 = 0.144 ]

     

  2. The level of nitrogen oxide in the exhaust of a new model of car when driven in city traffic is reported to have an approximately normal distribution with mean of 1.4 g/km and standard deviation of 0.19 g/km. We plan to take exhaust readings from 22 cars and compute their average NO exhaust level. What is the probability of observing a mean that is below 1g/km? (HINT: use the Central Limit Theorem to tell you what the sampling distribution of the mean should be, compute the z-scores for 1 g/km, then find the appropriate probability using Table 1).

    3. If you looked at the problems above you get the idea how to answer this question. Let X be the random variable representing the average NO exhaust level from 22 sample cars from a population with mean 1.4 and standard deviation 0.19. Then from the Central Limit Theorem, we are interested in P(X < 1) = P ( Z=(X-1.4)/(0.19/sqrt(22))<(1-1.4)/(0.19/sqrt(22)) = P(Z < -9.87 ) = 0.00000. That is, it is almost impossible that the NO exhaust average of the 22 sample cars will be less than 1 g/km (assuming that they come from the stated population.) ]
For students in community development, education and social services fields.
  1. (From Ott and Longnecker, p135, #4.16) A survey of a number of large corporations gave the following probability table for events related to the offering of a promotion that also involved a location transfer.
    		2.
    Married
    Promotion/Transfer
    Two-Career Marriage
    One-Career Marriage
    Unmarried
    Total
    Rejected
    .184
    .0555
    .0170
    .2565
    Accepted
    .276
    .3145
    .1530
    .7435
    Total
    .46
    .37
    .17

    Using this table compute the following:

    1. What is the probability that a professional (selected at random) would accept the promotion? [ .7435 ]
    2. What is the probability that a professional (selected at random) is part of a two-career marriage? [0.46]
    3. What is the conditional probability of accepting the promotion, given the professional is part of a two-career marriage? [P(A|B) = P(A and B)/P(B) Let A be the event of accepting the promotion and B be the event that the individual is part of a two-career marriage. P(B) = 0.46. P(A and B) = 0.276, P(A|B)=0.276/0.46 = 0.6. ]

     

  2. Based on the 1990 census, the numbers of hours per day that adults spend watching television is approximately normally distributed with a mean of 5 hours and a standard deviation of 1.3 hours. We plan to survey 50 adults and record the number of hours they watch television in a day. What is the probability that the average number of hours for our sample exceeds 6 hours per day? (HINT: use the Central Limit Theorem to tell you what the sampling distribution of the mean should be, compute the z-scores for 6 hours per day, then find the appropriate probability using Table 1).

    3. Let H be the random variable representing the average number of hours for 50 individuals sampled from a population with mean 5 and standard deviation 1.3. We want to know P(H > 6). From the Central Limit Theorem we know that the distribution of a standardized mean, Z= (H-5)/(1.3/sqrt(50)), will be a standard normal distribution. Thus P(H>6) = P(Z>(6-5)/(1.3/sqrt(50))) = P(Z>5.44) < 0.0001. Hence it is almost impossible for us to observe a average of 6 hours in a sample of 50 individuals who come from this population. If we did observe a mean of 6 this would strongly imply that the true population mean was not 5.