Math Statistics Week 5, Self Learning Quiz PDF

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This is a self unit quiz, doesn't count towards your grade...

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Math Statistics Week 5 Self Unit Approximately 85% of statistics students do their homework in time for it to be collected and graded. Let X be the number students that submit their homework in time out of a statistics class of 70 students. The following 4 questions refer to this X. (The answer may be rounded up to 3 decimal places of the actual value.) Question 1 Correct Mark 1.00 out of 1.00 Flag question

Question text The sample space of X is: Select one: a. The numbers between 0 and 1. b. The integers between (and including) 0 and 70. c. The collection {0, 1,...,85} d. The collection {"Submit on time", "Do not submit on time"}. Feedback The possible outcome of X, the number of students out of 70 that submit their homework, is an integer and the range of values starts at 0 (no one submits) and ends in 70 (all submit). Question 2 Correct Mark 1.00 out of 1.00 Flag question

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The probability that less than 60 of the 70 students will do their homework on time is: Answer: Feedback Less that 60 means 59 or less. The probability P(X ≤ 59) can be computed with the code: > pbinom(59,70,0.85) [1] 0.4842268 Question 3 Correct Mark 1.00 out of 1.00 Flag question

Question text The expectation of X is: Answer: Feedback The expectation of X is equal to n * p = 70 * 0.85 = 59.5 Question 4 Correct Mark 1.00 out of 1.00 Flag question

Question text The standard deviation of X is: Answer: Feedback The variance of X is n * p * (1-p) = 70 * 0.85 * 0.15 = 8.925. The standard deviation is the square root of the variance, namely 2.987474. The correct answer is: 2.987474. Information Flag question

Information text In Chapter 1 it was claimed that when tossing a fair coin 4 times it is quite likely to not obtain 2 heads and 2 tails. However, when tossing a fair coin 4,000 times one should expect to obtain number of tails in the range between 1940 and 2060. Let us compare the situation for 2 versus 2,000 coins. Let X be the number of heads when tossing a fair coin 2 times and

let Y be the number of heads when tossing a fair coin 2,000 times. (The answer may be rounded up to 3 decimal places of the actual value). Question 5 Correct Mark 1.00 out of 1.00 Flag question

Question text P( X = 1) is equal to Answer: Feedback The distribution of X is Binomial(2,0.5). The probability P(X = 1) can be computed with the code > dbinom(1,2,0.5) [1] 0.5 Question 6 Correct Mark 1.00 out of 1.00 Flag question

Question text P(X is not equal to 1) is Answer: Feedback If P(X = 1) = 0.5 then the complementary probability (X is not equal to 1) is equal to 1 - P(X = 1) = 1 - 0.5 = 0.5 Question 7 Correct Mark 1.00 out of 1.00 Flag question

Question text P(Y = 1,000) is equal to Answer: Feedback The distribution of Y is Binomial(2000,0.5). The probability P(Y = 1000) can be computed with the code > dbinom(1000,2000,0.5) [1] 0.01783901

Question 8 Correct Mark 1.00 out of 1.00 Flag question

Question text P(Y is not equal to 1,000) is Answer: Feedback If P(Y = 1000) = 0.01783901 then the complementary probability (Y is not equal to 1000) is equal to 1 - P(Y = 1000) = 1 - 0.01783901 = 0.982161 Question 9 Correct Mark 1.00 out of 1.00 Flag question

Question text P(940 ≤ Y ≤ 1,060) is equal to Answer: Feedback The probability P(940 ≤ Y ≤ 1,060) is equal to the difference between P(Y ≤ 1,060) and the probability P(Y < 940). The letter probability is equal to P(Y ≤ 939). This difference can be computed with the code: > pbinom(1060,2000,0.5) - pbinom(939,2000,0.5) [1] 0.9931974 Question 10 Correct Mark 1.00 out of 1.00 Flag question

Question text E(X) is equal to Answer: Feedback The expectation of X is equal to n * p = 2 * 0.5 = 1 Question 11 Correct

Mark 1.00 out of 1.00 Flag question

Question text The standard deviation of X is equal to Answer: Feedback The variance of X is equal to n * p * (1-p) = 2 * 0.5 * 0.5 = 0.5. The standard deviation is the square root of 0.5, which is equal to 0.7071068 Question 12 Correct Mark 1.00 out of 1.00 Flag question

Question text E(Y) is equal to Answer: Feedback The expectation of Y is equal to n * p = 2000 * 0.5 = 1000 Question 13 Correct Mark 1.00 out of 1.00 Flag question

Question text The standard deviation of Y is equal to Answer: Feedback The variance of Y is equal to n * p * (1-p) = 2000 * 0.5 * 0.5 = 500. The standard deviation is the square root of 500, which is equal to 22.36068 Information Flag question

Information text Meiosisis the process in which a diploid cell that contains two copies of the genetic material produces an haploid cell with only one copy (sperms and eggs). The resulting molecule of genetic material is linear molecule that is composed of consecutive segments: a segment that originated from one of the two copies followed by a segment from the other copy and vice versa. The border points between segments are called points of

crossover. The Haldane model for crossovers states that the number of crossovers between two loci on the genome has a Poisson(λ) distribution. Assume that the expected number of crossovers between two loci in a fixed period of time is 2.25.The next 3 questions refer to this model for crossovers. (The answer may be rounded up to 3 decimal places of the actual value.)

Question 14 Correct Mark 1.00 out of 1.00 Flag question

Question text The probability of obtaining exactly 4 crossovers between the two loci is Answer: Feedback The number of crossovers has a Poisson distribution with parameter λ = 2.25. The probability of exactly 4 crossovers can be computed with the code: > dpois(4,2.25) [1] 0.1125528 Question 15 Correct Mark 1.00 out of 1.00 Flag question

Question text The probability of obtaining at least 4 crossovers between the two loci is Answer: Feedback The number of crossovers has a Poisson distribution with parameter λ = 2.25. The probability of at least 4 crossovers can be computed as the difference between 1 and the probability of 3 or less crossovers. The computation can be conducted with the code: > 1 - ppois(3,2.25) [1] 0.1905669 Question 16 Correct Mark 1.00 out of 1.00 Flag question

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A recombination between two loci occurs if the number of crossovers is odd. The probability of recombination between the two loci is, approximately, equal to Answer (Compute the probability of recombination approximately using the function "dpois". Ignore odd values larger than 9) Feedback Values (not larger than 9) that lead to recombination are 1, 3, 5, 7, and 9. The probability of these values for the Poisson distribution with parameter that is equal to λ = 2.25 can be computed with the code: > sum(dpois(c(1,3,5,7,9),2.25)) [1] 0.4944251 Information Flag question

Information text The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between 0 and 17minutes, inclusive. The next 3 questions refer to this waiting time. (The answer may be rounded up to 3 decimal places of the actual value.) Question 17 Correct Mark 1.00 out of 1.00 Flag question

Question text The probability that a person waits fewer than 12.5 minutes is Answer: Feedback Let the X be the length of time the person waits. The distribution of X is Uniform(0,17). The probability P(X ≤ 12.5) can be computed with the code > punif(12.5,0,17) [1] 0.7352941 Question 18 Correct Mark 1.00 out of 1.00 Flag question

Question text The expectation of the waiting time is

Answer: Feedback Let the X be the length of time the person waits. The distribution of X is Uniform(0,17). The expectation of x is equal to (a+b)/2 = (0+17)/2 = 8.5 Question 19 Correct Mark 1.00 out of 1.00 Flag question

Question text The standard deviation of the waiting time is

Answer: Feedback Let the X be the length of time the person waits. The distribution of X is Uniform(0,17). The variance of x is equal to (b-a)^2/12 =(17-0)^2/12 = 24.08333. . The standard deviation is the square root of the variance and is equal to 4.907477 Information Flag question

Information text Let X be amount of time (in minutes) a postal clerk spends with his/her customer. Assume that X has an Exponential(λ) distribution and that E(X) = 7 minutes. The next 3 questions refer to this waiting time. (The answer may be rounded up to 3 decimal places of the actual value.) Question 20 Correct Mark 1.00 out of 1.00 Flag question

Question text The rate λ is equal to Answer: Feedback The expectation in the Exponential distribution is the reciprocal of the parameter λ.Consequently, the parameter λ is equal to the inverse of the expectation: λ = 1/E(X). The expectation E(X) = 7, hence λ = 1/7 = 0.1428571

Question 21 Correct Mark 1.00 out of 1.00 Flag question

Question text The probability that a clerk spends between four to five minutes with a randomly selected customer is Answer: Feedback Let the distribution of X be Exponential(1/7). The probability P(4 ≤ X ≤ 5) is equal to the difference between P(X ≤ 5) and the probability P(X < 4). The letter probability is equal to P(X ≤ 4), since the distribution is continuous. This difference can be computed with the code: > pexp(5,1/7)-pexp(4,1/7) [1] 0.07517646 Question 22 Correct Mark 1.00 out of 1.00 Flag question

Question text The probability that a clerk spends more than 10 minutes with a customer is Answer: Feedback Let the distribution of X be Exponential(1/7). The probability P(10 < X) is equal to the difference between 1 and the probability P(X ≤ 10). This difference can be computed with the code: > 1-pexp(10,1/7) [1] 0.2396510...