Quiz 21 10 Points Once Upon A Time I Had A Fast Food Lunch ✓ Solved

Once upon a time, I had a fast-food lunch with a mathematician colleague. I noticed a very strange behavior in him. I called it the Au-Burger Syndrome since it was discovered by me at a burger joint. Based on my unscientific survey, it is a rare but real malady inflicting 2% of mathematicians worldwide. Yours truly has recently discovered a screening test for this rare malady, and the finding has just been reported to the International Association of Insane Scientists (IAIS) for publication. Unfortunately, my esteemed colleagues who reviewed my submitted draft discovered that the reliability of this screening test is only 80%. What it means is that it gives a positive result, false positive, in 20% of the mathematicians tested even though they are not afflicted by this horribly-embarrassing malady. I have found an unsuspecting victim, oops, I mean subject, down the street. This good old mathematician is tested positive! What is the probability that he is actually inflicted by this rare disabling malady?

Most of us love Luzon mangoes, but hate buying those that are picked too early. Unfortunately, by waiting until the mangos are almost ripe to pick carries a risk of having 15% of the picked rot upon arrival at the packing facility. If the packing process is all done by machines without human inspection to pick out any rotten mangos, what would be the probability of having at most 2 rotten mangos packed in a box of 12?

We have 7 boys and 3 girls in our church choir. There is an upcoming concert in the local town hall. This performance team of 5 has to be picked randomly from the crew of 7 boys and 3 girls. a. What is the probability that all 3 girls are picked in this team of 5? b. What is the probability that none of the girls are picked in this team of 5? c. What is the probability that 2 of the girls are picked in this team of 5?

In this economically challenging time, yours truly, CEO of the Outrageous Products Enterprise, would like to make extra money to support his frequent filet-mignon -and- double-lobster-tail dinner habit. A promising enterprise is to mass-produce tourmaline wedding rings for brides. Based on my diligent research, I have found out that women's ring size normally distributed with a mean of 6.0, and a standard deviation of 1.0. I am going to order 5000 tourmaline wedding rings from my reliable Siberian source. They will manufacture ring sizes from 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, and 9.5. How many wedding rings should I order for each of the ring sizes should I order 5000 rings altogether?

A soda company wants to stimulate sales in this economic climate by giving customers a chance to win a small prize for every bottle of soda they buy. There is a 20% chance that a customer will find a picture of a dancing banana at the bottom of the cap upon opening up a bottle of soda. The customer can then redeem that bottle cap with this picture for a small prize. Now, if I buy a 6-pack of soda, what is the probability that I will win something, i.e., at least winning a single small prize?

When constructing a confidence interval for a population with a simple random sample selected from a normally distributed population with unknown σ, the Student t-distribution should be used. If the standard normal distribution is correctly used instead, how would the confidence interval be affected?

Below is a summary of the Quiz 1 for two sections of STAT 225 last spring. The questions and possible maximum scores are different in these two sections. We notice that Student A4 in Section A and Student B2 in Section B have the same numerical score. How do these two students stand relative to their own classes? And, hence, which student performed better? Explain your answer.

My brother wants to estimate the proportion of Canadians who own their house. What sample size should be obtained if he wants the estimate to be within 0.02 with 90% confidence if a. he uses an estimate of 0.675 from the Canadian Census Bureau? b. he does not use any prior estimates? But in solving this problem, you are actually using a form of "prior" estimate in the formula used. In this case, what is your "actual" prior estimate? Please explain.

An amusement park is considering the construction of an artificial cave to attract visitors. The proposed cave can only accommodate 36 visitors at one time. In order to give everyone a realistic feeling of the cave experience, the entire length of the cave would be chosen such that guests can barely stand upright for 98% of all the visitors. The mean height of American men is 70 inches with a standard deviation of 2.5 inches. An amusement park consultant proposed a height of the cave based on the 36-guest-at-a-time capacity. Construction will commence very soon. The park CEO has a second thought at the last minute, and asks yours truly if the proposed height is appropriate. What would be the proposed height of the amusement park consultant? And do you think that it is a good recommendation? If not, what should be the appropriate height? Why?

A department store manager has decided that dress code is necessary for team coherence. Team members are required to wear either blue shirts or red shirts. There are 9 men and 7 women in the team. On a particular day, 5 men wore blue shirts and 4 other wore red shirts, whereas 4 women wore blue shirts and 3 others wore red shirts. Apply the Addition Rule to determine the probability of finding men or blue shirts in the team.

It is an open secret that airlines overbook flights, but we have just learned that bookstores underbook textbooks in the good old days that we had to purchase textbooks. To make a long story short, once upon a time, our UMUC designated virtual bookstore, MBS Direct, routinely, as a matter of business practice, orders less textbooks than the amount requested by UMUC's Registrar's Office. That is what I have figured out. Simply put, MBS Direct has to "eat" the books if they are not sold.

Suppose you are the CEO of MBS Direct, and you want to perform a probability analysis. What would be the number of STAT 200 textbook bundles you would order so that you stay below 5% probability of having to back-order from Pearson Custom Publishing?

Is there an approximation method for Question 11? If so, please tackling Question 11 with the approximation method.

Paper For Above Instructions

In evaluating the probability that a mathematician testing positive for the “Au-Burger Syndrome” actually has the condition, we start with the following parameters. The prevalence of this syndrome is 2%, meaning of a sample of 100 mathematicians, 2 individuals are expected to truly have it. The screening test has a reliability of 80%, causing 20% of individuals who do not have the syndrome to test positive erroneously (false positive). Consequently, if a mathematician tests positive, the question arises: what is the probability that he actually is afflicted by the syndrome? Using Bayes' theorem, we can calculate this probability as follows:

Let A be the event that a mathematician has Au-Burger syndrome, and B the event that the test result is positive. We are interested in P(A|B), the probability that an individual has the syndrome given that they tested positive. We can calculate it using the formula:

P(A|B) = [P(B|A) * P(A)] / P(B)

Where:

  • P(B|A) = Probability of testing positive given that the individual has the syndrome = 80% or 0.8

  • P(A) = Probability of having the syndrome = 0.02

  • P(B) = Total probability of testing positive, which can be found using the law of total probability:

P(B) = P(B|A) P(A) + P(B|¬A) P(¬A)

Where P(B|¬A) is the probability of testing positive given that the individual does not have the syndrome, which equals 20% or 0.2, and P(¬A) is the probability of not having the syndrome which is 1 - P(A) = 0.98.

Now calculating P(B):

P(B) = (0.8 0.02) + (0.2 0.98) = 0.016 + 0.196 = 0.212

Now substituting these values into the equation for P(A|B):

P(A|B) = (0.8 * 0.02) / 0.212 = 0.016 / 0.212 ≈ 0.0755

Thus, the probability that the mathematician who tested positive actually has the syndrome is approximately 7.55%.

This example illustrates how the accuracy of a test can impact the interpretation of its results, especially when the condition being tested for is quite rare.

Next, we analyze the probability of having at most 2 rotten mangos in a box of 12, given that 15% of the picked mangos rot. This situation can be modeled using the binomial distribution, where n=12 (the size of the box) and p=0.15 (the probability of a mango rotting). The probability mass function for a binomial distribution is given as:

P(X=k) = C(n, k) (p^k) ((1-p)^(n-k))

Where C(n, k) is the binomial coefficient representing the number of ways to choose k successes in n trials. Here, we seek:

P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)

Calculating these, we have:

P(X=0) = C(12, 0) (0.15^0) (0.85^12) ≈ 0.1575

P(X=1) = C(12, 1) (0.15^1) (0.85^11) ≈ 0.2416

P(X=2) = C(12, 2) (0.15^2) (0.85^10) ≈ 0.2277

Thus, we add the probabilities:

P(X ≤ 2) ≈ 0.1575 + 0.2416 + 0.2277 ≈ 0.6268

This indicates a ~62.68% probability of having at most 2 rotten mangos packed in a box of 12.

In terms of selecting a performance team of 5 from a choir of 7 boys and 3 girls, we can similarly use combinatorics related to combinations. The total number of ways to choose 5 from 10 (7 boys + 3 girls) is given by C(10, 5) = 252. To find the probability of selecting all 3 girls in the team, it follows that we must also select 2 boys:

Probability(all 3 girls) = C(3, 3) C(7, 2) / C(10, 5) = 1 21 / 252 = 0.0833.

For the probability that none of the girls are picked, we only select from boys:

Probability(no girls) = C(7, 5) / C(10, 5) = 21 / 252 = 0.0833.

Finally, for the probability that exactly 2 girls are selected:

Probability(2 girls) = C(3, 2) C(7, 3) / C(10, 5) = 3 35 / 252 = 0.4167.

Moving to the analysis for wedding rings, where we expect a mean ring size of 6.0 and a standard deviation of 1.0. Given that size distributions are assumed normal, we’ll need to outline what proportions of rings to manufacture at each size. The standard normal variable Z can be applied to find the respective percentages for different size ranges. With 5000 rings, and adopting an approximate cut-off method for sizes via Z-tables, we can allocate sizes proportionally according to the calculations of ring sizes that fall within standard deviations of the mean.

For the probability of winning at least one prize from a 6-pack of soda where each has a win chance of 20%, we can delineate this as:

P(winning at least one) = 1 - P(winning none) = 1 - (0.8^6) ≈ 0.2621.

As we engage with confidence intervals, utilizing the t-distribution rather than the standard normal distribution affects the margin of error consequently resulting in potentially wider intervals, as the sample size diminishes and degrees of freedom come into play with the t-values being greater than those of the Z-values.

Evaluating student performances across two sections focuses on standard deviations and z-scores relative to mean scores for each section, necessitating comparison across distributions to ascertain who performed better simply by standardization.

Finally, with the underbooking malpractice of textbooks to meet a statistical threshold while minimizing excess inventory, utilizing probability functions and anticipated enrollment percentages permits calculation of needs adequately.

References

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