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QUESTION

Need my answers checked for this problem. I do not have a probability distrabution to work with only what I found on my calculator as noted.

Need my answers checked for this problem. I do not have a probability distrabution to work with only what I found on my calculator as noted. 

Specifically needing help forming my answers for number #5 and #7

A manager of restaurant knows that 30% of his customers are professional truckers. The manager surveys the next 12

customers who come into the restaurant to determine if they are professional truck drivers or not.

1. Does this situation meet the binomial requirements? Why or why not?

Address the five conditions of a binomial distribution in the textbook.

Yes.

Reason 1. Fixed number of trials.

Reason 2. There are 2 possible outcomes for each trial. (Truck Driver or Not a Truck Driver).

Reason 3. Probability of success is the same for each trial.

Reason 4. The trials are independent.

Reason 5. The random variable “X” represents the number of people who are found to in fact be truck drivers.

2. What is the random variable in this problem? (i.e., what does X represent and what values can it have?)

X represents a person being asked if they are a truck driver or not.

3. What is the probability that exactly 4 are truck drivers?

n=12 P=.30 x=4

binompdf(12,0.3,4)

.2311396961 or .2311

4. What is the probability that 5 or fewer are truck drivers?

n=12 P=.30 x=5

binomcdf(12,0.3,5)

.882151261 or .88

5. What is the probability that at least 4 are truck drivers?

P(Between 4 and 12, inclusive) = Addition rule for mutually exclusive events.

P(1) = .0711837627

P(2) = .1677902979

P(3) = .2397004256

P(4) = .2311396961

P(5) = .1584957916

P(6) = .0792478958

P(7) = .0291114719

P(8) = .0077977157

P(9) = .0014852792

P(10) = 1.91E-4

P(11) = 1.49E-5

P(12) = 5.31E-7

6. What is the probability that at most 4 are truck drivers?

P(between 1 and 4, inclusive) = P(Less than or equal to 4) - P(Equal to 0)

n=12 p=.30 x=4

binomcdf(12,0.30,4)

.7236554696 or .72

n=12 p=.30 x=0

binompdf(12,.30,0)

.0138412872 or .01

Answer .71 - .01 = .71

7. What is the mean number of customers out of a sample of 12 that you would expect to be truck drivers? What is the

standard deviation of the distribution?

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