Solution:- ( I Got the answer for number one already) just need two and three 1)

Question

Solution:- ( I Got the answer for number one already) just need two and three

1) The probability of getting at most 15 questions correct is 0.8852.

p = 0.50

n = 25

x = 15

By applying binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x < 15) = 0.8852

1. (35 pts) A student is going to take a true/false quiz with 25 questions. If the student randomly selects an answer for each question, find the probability that the student’s score is 60% or less (in other words, find the probability of getting at most 15 questions correct). HINT: Model using the binomial distribution to first determine n, , and the values of X. Then use MINITAB to find the required probability. When you provide your answer, be sure to state the values of n and that you used in addition to the probability requested. Also, see the hint at the end of this assignment. . ( I Got the answer for number one already) just need two and three

2. (30 pts) A standard roulette wheel in the U. S. has the numbers 1 through 36 plus 0 and 00. Of the numbers 1 through 36, 18 are colored red and 18 are colored black. The numbers 0 and 00 are colored green. If a person observes 20 consecutive spins of the wheel, what is the probability that the number of times that the color red occurs is more than 10? You can assume that results of separate spins are independent so that this probability can be modeled using the binomial distribution. HINT: Model using the binomial distribution to first determine n, , and the values of X. Then use MINITAB to find the required probability. When you provide your answer, be sure to state the values of n and that you used as well as the probability requested. Also, see the hint at the end of this page.

3. A student group at a university raises funds by undertaking small projects. The group’s next project requires a decision concerning the number of copies of a brochure to be printed for homecoming weekend. The printing order must be a multiple of 500. The size of the homecoming crowd is uncertain because it depends on weather conditions. The following payoff table has been prepared (payoffs are net profits in dollars).

Printing Order Size of Crowd 1000 1500 2000 2500 3000 Above average 200 300 400 500 450 Average 200 300 250 200 150 Below Average 200 150 100 50 0

a) (10 pts) Develop the regret table from this payoff table and use it to determine the best choice for the number of brochures to print based on the criterion of minimax regret.

b) (15 pts) The group estimates that the probability of an above average crowd is .20, the probability of an average crowd is .60, and the probability of a below average crowd is .20. Using the criterion of mean profit (EMV), which is the best brochure size to print? Justify your answer by showing EMV for each of the five printing order sizes.

. c) (10 pts) The group estimates that the probability of an above average crowd is .20, the probability of an average crowd is .60, and the probability of a below average crowd is .20. Using the criterion of mean regret (EOL), which is the best brochure size to print? Justify your answer by showing EOL for each of the five printing order sizes

. HINT FOR PROBLEMS 1 and 2: These questions are easier to answer using cumulative probabilities. Remember that a cumulative probability is the probability of all values less than or equal to X.

Here are some example of how that can be used. I’ve used 100 as an arbitrary value for X to illustrate the idea. P(X 100) is the cumulative probability with x equal to 100. P(X > 100) = 1 – P(X 100) P( X 100 ) = 1 – P( X 99), where P(X 99) is the cumulative probability with x equal to 99. P( X < 100) = P( X 99)

Explanation / Answer

BIONOMIAL DISTRIBUTION

pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k

where

k = number of successes in trials   

n = is the number of independent trials   

p = probability of success on each trial

p = the number of times that the color red occurs = 18/38 = 0.47368

I.

mean = np

where

n = total number of repetitions experiment is excueted

p = success probability

mean = 20 * 0.47368

= 9.4736

II.

variance = npq

where

n = total number of repetitions experiment is excueted

p = success probability

q = failure probability

variance = 20 * 0.47368 * 0.52632

= 4.98615

III.

standard deviation = sqrt( variance ) = sqrt(4.98615)

=2.23297

P( X < = 10) = P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)

= ( 20 10 ) * 0.47368^10 * ( 1- 0.47368 ) ^10 + ( 20 9 ) * 0.47368^9 * ( 1- 0.47368 ) ^11 + ( 20 8 ) * 0.47368^8 * ( 1- 0.47368 ) ^12 + ( 20 7 ) * 0.47368^7 * ( 1- 0.47368 ) ^13 + ( 20 6 ) * 0.47368^6 * ( 1- 0.47368 ) ^14 + ( 20 5 ) * 0.47368^5 * ( 1- 0.47368 ) ^15 + ( 20 4 ) * 0.47368^4 * ( 1- 0.47368 ) ^16 + ( 20 3 ) * 0.47368^3 * ( 1- 0.47368 ) ^17 + ( 20 2 ) * 0.47368^2 * ( 1- 0.47368 ) ^18 + ( 20 1 ) * 0.47368^1 * ( 1- 0.47368 ) ^19 + ( 20 0 ) * 0.47368^0 * ( 1- 0.47368 ) ^20

= 0.67767

P( X > 10) = 1 - P ( X <=10) = 1 -0.67767 = 0.32233

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