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Math 361 Practice Exam 2 (Use this information for questions 1 – 3) At the end of a production run manufacturing rubber gaskets, items are sampled at random and inspected to determine if the item is Acceptable (A), or Defective (D). Suppose it is planned to select two items and determine if each is either A or D. It is estimated that about 12% of the items produced are defective. 1.) What is the Sample Space, S, of the outcomes? S = {AA, AD, DA, DD} 2.) What is the set of outcomes that describes the event that at least one of the items selected is defective? {AD, DA, DD} 3.) What is the probability that at least one item selected is defective? P(X ≥ 1) = 0.88*0.12 + 0.12*0.88 + 0.12*0.12 = 0.2256 4.) Suppose that S is a sample space. P(S) = 1 (The probability of getting one of the possible outcomes is equal to 1 since something has to come out of the experiment.) Use the following to answer questions 5-11: The probability distribution of random variable, X, is defined as follows: X Probability -5 0.4 5 0.2 6 0.1 5.) State the conditions for this to be a valid probability model: a.) Probabilities for each outcome must be between 0 and 1. b.) The probabilities of all possible outcomes must add to one. 6.) Find the expected value of the random variable: E(x) = -5*0.4 + 5*.2 + 6*0.1 + 0*.1 + 2*0.2 = 0 0 0.1 2 0.2 7.) Given that the variance of this distribution is 90, determine the mean and standard deviation of the random variable Y where Y = 3 + 2x E(Y) = E(3 + 2x) = E(3) + 2*E(x) = 3 + 2* 0 = 3 Var(Y) = E[ (Y – E(Y))2) = E[ (Y – 3)2] = E[ Y2 – 6Y + 9] = E(Y^2) – 6*E(Y) + E(9) = E[ (3 + 2x)2) ] – 6*3 + 9 = E[ 9 + 12X + 4x2 ] -9 = E(9) + 12*E(X) + 4*E(X2) – 9 = 9 + 12*0 + 4*E(X2) – 9 = 4*E(X2) Var(X) = E[ (X – E(x))2] = E(x2) = 90 So: Var(Y) = 4*90 = 360 Standard deviation of Y = 18.97366596 8.) Suppose you draw two numbers at random from this distribution. What is the mean and standard deviation of the sum of these two numbers? µ = E(X + X) = E(X) + E(X) = 0 + 0 = 0 σ2 = 90, So σ2x+x = 90 + 90 = 180 σ= 9.) Find P(X = 6)? 0.1 10.) Find P(X < 5)? 0.4 + 0.1 + 0.2 = 0.7 11.) What is the median of X? 0 (Remember, median is the lowest category including 0.5 probability. Write them in order to see this.) 12. Suppose X is a random variable with the given distribution. What is P(1 ≤ X ≤ 2)? = Area under the curve = ½ * Base * Height = ½*1*1/2= ¼ 13.) Suppose there are three balls in a box. On one of the balls is the number 1, on another is the number 2, and on the third is the number 3. You select two balls at random and without replacement from the box and note the two numbers observed. The sample space S consists of the three equally likely outcomes {(1, 2), (1, 3), (2, 3)} (disregarding order). Let X be the sum of the two balls selected. a.) Write the probability distribution for X X Probability of X 3 1/3 4 1/3 5 1/3 b.) What is the probability that the sum is at least 4. 2/3 c.) What is the mean of x? 3*1/3 + 4*1/3 + 5*1/3 = 12/3 = 4 Use the following information for questions 14 and 15 The weight of medium-size tomatoes selected at random from a bin at the local supermarket is a random variable with mean µ = 12 oz and standard deviation σ = 2 oz 14.) Suppose we pick 4 tomatoes from the bin at random. a.) How much do you expect the weight of the bag of tomatoes to be? 4*12 = 48oz b.) Find the standard deviation associated with the weight of the bag of tomatoes. Variance = 4 for a single tomato. Var(4*X) = 16*var(x) = 32. Standard Deviation = oz 15.) Suppose you only choose two tomatoes and you create the variable Y which is the difference in weight between the larger and smaller tomato. a.) What is the mean of Y? E(X – Y) = E(X) – E(Y) = 0 – 0 = 0 b.) What is the standard deviation of Y? Var(X – Y) = 4 + 4 = 8 so SD = Use the following for questions 18 - 21 In a large city, 72% of the people are known to own a cell phone, 38% are known to own a pager, and 29% own both a cell phone and a pager. Let A be the event that they own a cell phone and B be the event that they own a pager. 18.) Fill in a venn diagram representing this situation. 19.) Suppose an individual is drawn at random from this population. Find the following probabilities. a.) = 0.81 b.) = 0.29 c.) d.) = 0.71 = 43/(43 + 19) = 0.693548 20.) Are these two events mutually exclusive? Explain. No, there is an overlap between the events. 21.) Are these two events independent? Explain. P(A) = 0.72, P(A | B) = 29/38) = 0.763158 Because P(A) ≠ P(A|B) These two events are not independent. (They are pretty close though, since the probability doesn’t change much). Use the following for questions 22 – 24: Bob has recently been hired by a shop downtown to help customers with various computer related problems. Lately, two different viruses have been bugging many customers—virus Dummy and virus Smarty. It is estimated that about 65% of the customers with virus problems are bothered by virus Dummy and the remaining 35% by virus Smarty. If the computer is infected by virus Dummy, Bob has a 90% chance of fixing the problem. However, if the computer is infected by the virus Smarty, this chance is only 70%. 22.) Complete the following tree diagram representing this situation. 23.) If a computer is brought in by a customer, what is the probability that Bob will be able to fix it? P(Fix) = 0.585 + 0.245 = 0.83 24.) If Bob was able to fix a computer, what is the probability that it was infected by virus Dummy? Prob(Dummy | Fix) = 0.585/(0.585+0.245) = 0.704819 25.) If 12 students are to be selected, what is the probability that more than 7 students in the sample voted for the democratic student party if, in a recent poll, 45% claimed to be democrat? First, they either are or are not, so this is Bernoullii trials. And, since we are looking for r successes in N trials, this is binomial (pdf). Binomialpdf (12,0.45,7) = 0.148945 Use the following to answer questions 26 – 30: Chromosome defect A occurs in only one out of 200 adult males. A random sample of 1000 adult males is selected. Let the random variable X represent the number of males in the sample who have this chromosome defect. 26.) How many individuals from the sample do we expect to have this defect? P = 1/200 = 0.005 and 0.005*1000 = 5 27.) What is the standard deviation on the number we expect to have this defect? Standard Deviation of a binomial = = = 2.23047 28.) What is the probability that the first individual selected with this defect occurs on the 200th person sampled? Geometric: 0.995*0.995*…*0.995*0.005 = 0.995199*0.005 = 0.001844 29.) What is the probability that you get more than 6 people in the sample with this defect? P(X > 6) = 1- P(X ≤ 6) = 1 – binomialcdf(1000, 0.005, 6) = 1 – 0.76255 = 0.2374492951 30.) What is the probability that you find 5 or less people in the sample with this defect? binomialcdf(1000, 0.005, 6) = 0.6159610219 Use the following information to answer questions 31 – 35: Suppose that you are considering IQ scores, which are normally distributed with a mean of 100 and a standard deviation of 15. Determine the following. 31.) When selecting a single individual at random, what is the probability that they have a score of 80 or less? Z = (80 – 100)/15 = -1.33. Prob = 0.091211 32.) When selecting a single individual at random, what is the probability that they have a score between 90 and 130? Z1 = (130 - 100)/15 = 2.00 Prob = 0.977249938 Z2 = (90 – 100)/15 = -0.67 Prob = 0.252492467 Prob in the middle = 0.977249938 – 0.252492467 = 0.724757471 33.) When selecting a single individual at random, what IQ score is necessary in order for that individual to be in the top 8% of the population? Prob of 0.08 on the right means Prob of 0.92 on the left. This means Z = 1.40507 In this case: 1.40507 = (x – 100)/15 so X = 121.076 34.) When selecting a single individual at random, what is the probability that they have a score of 145 or higher? P(X >= 145) = 1 – P(X < 145) Z = (145 – 100) / 15 = 3. 0.0013499672 35.) Suppose that we select 9 individuals randomly from the population. What is the probability that the mean IQ from this group is greater than 105? Z = (105 – 100)/(15/sqrt(9)) P = 0.1586552596 36.) Suppose that we select 25 individuals randomly from the population. What is the probability that the mean IQ from this group is greater than 105? Z = (105 – 100)/(15/sqrt(25)) P = 0.0477903304