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probability distributions assignment solution

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Finding Probabilities from Distributions and Study Cases

The solutions below are based on finding probabilities from probability distributions like the normal distribution. Others are concerned with calculating the probabilities of occurrence of various events in given cases without distributions. None of the solutions is wrong, and each of them is original and neatly presented.

Normal Distribution Mean and Probabilities

To make sure that sandwiches from the company are actually 12 inches long, a team conducted a survey in which they measured the length of 20 sandwiches from 10 different Subways. The average lengths of the sandwiches are 11.8 inches with a sample standard deviation value of 0.2 inches. Assuming that this survey can be considered a simple random sample and the data follows a normal distribution, what is the probability that, the length of a sandwich is between (11.9-11.95)?

Solution

Sandwich mean = 20

    Sandwich SD = 11.8

    Sandwich is normally distributed with (mean, SD)

Probability = P(11.9 < sandwich < 11.95)

                 = (11.9-20)/11.8< sum <(11.95-20)/11.8

                  = P(-0.6864 < z < -0.6822)

                  = 0.0013258.

Randomized controlled trials are used to determine the effectiveness of medical procedures — a timely topic during the current pandemic. Suppose that 200 people were recruited for a double-blinded medical trial for a COVID 19 Vaccine. Furthermore, suppose that of the 100 people who were assigned to the placebo group 10 got Covid 19, and of the 100 people that were assigned to the treatment group, 3 got Covid 19.

1. Assume from this data that the true probability of a random person getting Covid 19 is 1/10. What is the probability that the treatment has the same effect as the placebo and that the 3 out of 100 people who caught Covid was just due to randomness?

Solution

P (treatment placebo) = 100/200 = 1/2

   P (A= covid positive) = 10/100 = 1/10

 P(B = treatment positive) = 3/100

Since the effect is due to randomness, the probability that the treatment has the same effect

 P (same effect) = 1/10* 3/100

                          = 0.003

2. Let’s now instead assume that an unusually large number of people got Covid in the control group just by randomness. In particular, let's assume that the true probability of a random person getting Covid 19 is 6/100. Using these assumptions compute the probability that the treatment has the same effect as the placebo and that the 3 out of 100 people who caught Covid was just due to randomness.

Solution

since the probability for having covid increase to 6/100

Also, the effect is due to randomness.

Hence, the probability equals 1/10* 6/100

                           = 0.006

Using Probability to Make Decisions

For the following problems assume that you are on the American game show Let's Make a Deal with Monty Hall. Due to the growing popularity of the show thanks to "The Monty Hall Problem", the producers decided to add a couple more games of chance to the show. Assume that you are playing the show and it is your goal to maximize your expected winnings.

Monty Hall presents you with 2 games of chance. In the first, game 1 (G1), a fair coin is tossed and if it lands heads you win and if it lands tails you lose. In the second, game 2 (G2), S dice are thrown once, and if all 5 lands on the same number you win (this is also known as a Yahtzee). Monty Hall says that he will give you 1 million dollars if you can beat him in 2 of these games in a row. There is however catch, you must choose between playing him in either the first game, the second game, and then the first game again i.e. G1G2G1, or the second game than the first game, and then the second game again G2G1G2 Which order should you choose? NOTE: you must use the probability theory learned in the class to answer this question.

Solution

Probability of win (G1) = 1/2

Probability of win (G2) = 5C1 * ½^5

                                       = 5/32

P(G1 G2 G1) = 1/2* 5/32* 1/2

                        = 5/128

P(G2 G1 G2) = 5/32* 1/2* 5/32

         = 25/2048

In conclusion, the first order is desired which is G1G2G1 which has the higher probability among the two.