A. The table below contains information on the number of daily emergency service

Question

A. The table below contains information on the number of daily emergency service calls received by the volunteer ambulance service of Youngstown for the last 50 days: 22 days of which 2 emergency calls were received, 9 days of which 3 emergency calls were received, 8 days of which no emergency calls were received, etc. Number of NuallsDays Service Calls Number of 10 4 Total 50 1. What is the probability distribution of X? 2. Is this an example of a discrete or continuous probability distribution? Explain. 3. Is this a valid probability distribution? Explain. 4. What is the most likely number of calls that the agency might receive in a given day? 5. How many calls should the agency expect to receive in a given day? B. Nine percent of undergraduate students carry credit card balances greater than $7000 (Reader's Digest, July 2002). Suppose we randomly select 10 students from YSU and interview them about credit card usage. 1. Why is the selection of the 10 students a binomial experiment? Please be specific. 2. What is the probability that two of the students will have a credit card balance greater than $7000? 3. What is the probability that at least three students will have a credit card balance greater than $7000? 4. Consider the outcomes in parts 'b' and'e', which one is more likely and why? 5. Assuming the total enrollment this semester is 15,000, how many students would you expect to carry credit card balances greater than $7000?

Explanation / Answer

As per the chegg policy, we are advised to do one question at a time so i am attempting A.

A) 1. Probability distribution of X will be:

2. This is an example of discrete probability distribution because the values of x are whole numbers.

3. Yes, this is a valid probability distribution because all the probabilities are between 0 and 1 and the sum of all the probabilities is 1.

4. Most likely number of calls = Calls occuring with greatest frequency = 2 calls

5. Expected number of calls = 0(0.16) + 1(0.20) + 2(0.44) + 3(0.18) + 4(0.02) = 1.7

x p(x) 0 8/50 = 0.16 1 10/50 = 0.20 2 22/50 = 0.44 3 9/50 = 0.18 4 1/50 = 0.02

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