# Probability Distribution

1.The sales of Lexus automobiles in the Detroit area follow a Poisson distribution with a mean of 3 per day.

What is the probability that for five consecutive days at least one Lexus is sold?

2. An auditor for Health Maintenance Services of Georgia reports 40 percent of policyholders 55 years or older submit a claim during the year. Fifteen policyholders are randomly selected for company records.

(a) How many of the policyholders would you expect to have filed a claim within the last year?

(b) What is the probability that 10 of the selected policyholders submitted a claim last year? (Round your answer to 4 decimal places.)

(c) What is the probability that 10 or more of the selected policyholders submitted a claim last year? (Round your answer to 4 decimal places.)

(d) What is the probability that more than 10 of the selected policyholders submitted a claim last year? (Round your answer to 4 decimal places.)

3. According to the "January theory," if the stock market is up for the month of January, it will be up for the year. If it is down in January, it will be down for the year. According to an article in The Wall Street Journal, this theory held for 27 out of the last 34 years. Suppose there is no truth to this theory; that is, the probability it is either up or down is 0.5.

What is the probability this could occur by chance?

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1. The sales of Lexus automobiles in the Detroit area follow a Poisson distribution with a mean of 3 per day.

What is the probability that for five consecutive days at least one Lexus is sold?

Let X denotes the sales of Lexus automobiles in the Detroit area. It is given that, X follows a Poisson distribution with a mean of 3 per day. Thus, the probability distribution of X is given by,

, x = 0,1,...

Now, the probability that at least one Lexus is sold per day is give by,

P(X ≥ 1) = 1 - P(X = 0) = 1 - P(0) = 1 -

= 1 - 0.0498

= 0.9502

Thus, the probability that for five consecutive days at least one Lexus is sold

= (0.9502)^5

= 0.7746

2. An auditor for Health ...

#### Solution Summary

The solution discusses the determination of various probabilities using probability distributions.