Problem 3 Historical data indicates that only 30% of cable customers are willing

Question

Problem 3 Historical data indicates that only 30% of cable customers are willing to switch companies. If a binomial process is assumed, then in a sample of 10 cable customers, what is the probability that exactly 3 customers would be willing to switch their cable? Problem Historical data indicates that only 30% of cable customers are willing to switch companies. If a binomial process is assumed, then in a sample of 10 cable customers, what is the probability that between 3 and 5 (inclusive) customers are willing to switch companies? Problem 5 The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 400 and a standard deviation of 50. What is the probability that a student uses fewer than 350 minutes? Problem 6 The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 400 and a standard deviation of 50. What is the probability that a student uses more than 350 minutes? Problem? The time required to travel downtown at 10 a.m. on Monday morning is known to be normally distributed with a mean of 60 minutes and a standard deviation of 10 minutes. What is the probability that it will take less than 50 minutes? Problem 8 Through a telephone survey, a low-interest bank credit card is offered to 400 housecholds. The responses are as in the following table. Income $50,000 50 Income > $50,000 30 120 Accept offer Reject offer 200 Develop a joint probability table and show the marginal probabilities. a. What is the probability of a household whose income exceeds $50,000 and who rejects the offer? b. If income is less than or equal $50,000, what is the probability the offer will be accepted? c. d. If the offer is accepted, what is the probability that income exceeds $50,0002

Explanation / Answer

As multiple questions are mentioned, only the first one will be answered.

Problem 3:

For a binomial process (X) with n = 10 and p = 30%,

P(X=3) = nC3 * p3 * (1-p)10-3 = 10C3 * 0.33 * 0.77 = 0.2668

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