Mission College Quantity Business Analysis and Optimal Decision Strategy Worksheet

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Please check out attachment below for question details. Total 4 questions

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©Dr. Kezban Yagci Sokat BUS2 190 Spring 2021 BUS2 190 Quantitative Business Analysis Spring 2021 Homework 11 1. Suppose that you own a nail salon. The number of clients you serve each week is a random variable, C, with a binomial distribution. Using the following information, calculate the quantities below. 𝑃(𝐶 ≤ 65) = .96, 𝑃(𝐶 ≤ 64) = .90, 𝑃(𝐶 ≤ 55) = .86, 𝑃(𝐶 ≤ 54) = .84 𝑃(𝐶 ≤ 53) = .81, 𝑃(𝐶 ≤ 37) = .64, 𝑃(𝐶 ≤ 36) = .60, 𝑃(𝐶 ≤ 35) = .55 (a) (b) (c) (d) 𝑃(𝐶 ≥ 54) 𝑃(36 ≤ 𝑋 ≤ 54) 𝑃(𝐶 ≤ 65 | 𝐶 ≥ 37) 𝑃(𝐶 = 55) 2. You are a fundraiser for SJSU. To raise funds, you can run either a telephone campaign or an email campaign. If you run a telephone campaign, there’s a 40% chance that you will raise $100,000 and a 60% chance that you’ll raise $150,000.An email campaign will raise $90,000 with probability 30% and $140,000 with probability 70%. a) Use a decision tree to determine the best type of campaign to run. b) Draw a risk profile for the optimal decision. c) Right now there is a 70% chance that an email campaign will raise $140,000. How much higher or lower would this probability need to be in order for you to switch the type of campaign that you would run? d) Right now the most that a telephone campaign would raise is $150,000. How much higher or lower would this amount have to be in order for you to switch the type of campaign that you would run? 3. You own a shoe store and are deciding which shoes to stock for the upcoming season. You can either stock designer shoes, midrange shoes, or inexpensive shoes. If you stock designer shoes, you can stock either American designers or European designers. Based on which shoes you stock, your profits will be as follows: 1 ©Dr. Kezban Yagci Sokat BUS2 190 Spring 2021 (For example, if you stock shoes by European Designers there is a 50% chance that your profits will be $80,000.) (a) Use a decision tree to determine which shoes to stock in order to maximize expected profit. (b) Draw a risk profile for the optimal decision. (c) The probability of earning $50,000 when you stock inexpensive shoes is .3. How high would this probability need to be for you to decide to stock inexpensive shoes? 4. Mary is organizing an outdoor art show which will take place on August 15. The earnings from the show will depend heavily on the weather. If it rains on August 15, the show will lose $20,000. If it is sunny on August 15, the show will earn $15,000. Historically, the likelihood of it raining on any given day in mid-August is 27%. Suppose that today is July 31. Mary has the option of canceling the show by the end of the day on July 31, but if she does so, she will then lose her $1000 deposit on the facilities. (a) What is Mary’s optimal decision strategy? (b) Suppose the July 31 deadline has passed and Mary hasn’t cancelled the show. Mary can still cancel the show on August 14, but if she does so she must pay a fee of $10,000. The advantage of waiting until August 14 is that she can listen to the weather forecast for the next day on the local news station. According to station records, the weather was forecast to be sunny 90% of the days in mid-August in previous years. Also, when the weather was forecast to be sunny, it actually turned out to be sunny 80% of the time. When the weather was forecast to be rainy, it turned out to be rainy 90% of the time. What is Mary’s optimal decision strategy in this case? 2
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Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.This is for question 1

Question 1 Solution
1.)
a.) P(C≥54)
P(C ≤ 65) = .96 Applicable because the number of customers should be greater than or equal to 54
P(65≥54)= 1-P(65≤53)=1-0.0001=0.9999
P(C ≤ 64) = .90 Applicable because the number of customers should be greater than or equal to 54
P(64≥54)= 1-P(64≤53)=1- 0.05157= 0.94843
P(C ≤ 55) = .86 Applicable because the number of customers should be greater than or equal to 54
P(55≥54)= 1-P(55≤53)=1- 0.99751= 0.00249
P(C ≤ 54) = .84 Applicable because the number of customers should be greater than or equal to 54
P(54≥54)= 1-P(54≤53)=1- 0.99992= 0.00002
b.) P(36 ≤ C ≤ 54)
or can be interpreted as P(C≥36) and P(C ≤54)
P(C≤ 54) = .84 Applicable because 54 is between or equal to 36 and/or 54

P(C≥36)= 1-P(54≤35)=1-0.00046=0.99954
P(C≤54)= 1
The answer is between 0.99954 and 1
P(C≤ 53) = .81 Applicable because 53 is between or equal to 36 and/or 54

P(C≥36)= 1-P(53≤35)= 1-0.00716= 0.99284
P(C≤54)= P(53≤54) or P (54≥53)=1-P(54≤52)=1-0.99984= 0.00016
The answer is between 0.00016 and 0.99284
P(C≤ 37) = .64 Applicable because 37 is between or equal to 36 and/or 54

P(C≥36)= 1-P(37≤35)= 1-1=0
P(54≥37)=1-P(54≤36)=1-0.7054=0.2946
The answer is between 0 (the mean) and 0.2946

P(C ≤ 36) = .60 Applicable because 36 is between or equal to 36 and/or 54

P(C≥36)= 1-P(36≤35)=1-1=0
P(C≤54)=P(54≥36)= 1-P(54≤35)= 1-0.80465=0.19535
The answer is between 0 (the mean) and 0.19535
c) P(C≤ 65 | C ≥ 37) only on at a certain region
P(C ≤ 65) = .96 Applicable because it’s in the range between 37 and 65
P(65≤65)= 1
P(C≥37)=P(65>37)=1-P(65≤36)=0
The answer is between 0 (the mean) and 1
P(C ≤ 64) = .90 Applicable because it’s in the range between 37 and 65
P(64≤65) or P(65≥64)=1-P(65≤63)=1-0.99128= 0.00872
P(C≥37)= 1-P(64≤36)=1-0=1
The answer is between 0.00872 and 1
P(C ≤ 55) = .86 Applicable because it’s in the range between 37 and 65
P(55≤65) or P(65≥55)= 1-P(65≤54)=1-0.40666=0.59334
P(C≥37)= 1-P(55≤36)=...


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