SOLUTION: do i take .80/0, .80/1, .80/2, .80/3, .80/4 to find my probabilty???? If i am in the wrong thought process please help set me straight.
First fill in this table (Round to 3 dec
Algebra ->
Probability-and-statistics
-> SOLUTION: do i take .80/0, .80/1, .80/2, .80/3, .80/4 to find my probabilty???? If i am in the wrong thought process please help set me straight.
First fill in this table (Round to 3 dec
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Question 423430: do i take .80/0, .80/1, .80/2, .80/3, .80/4 to find my probabilty???? If i am in the wrong thought process please help set me straight.
First fill in this table (Round to 3 decimals)
BINOMIAL Probability Distribution
N=4 P=.80
Number of successes(x) probability
0 ?
1 ?
2 ?
3 ?
4 ?
2a) Mean= expected value for the # of sales= ____
Variance = _________
Use the table to answer the following questions:
2b) What is the probability that atleast1 policy is sold?
2c) What is the probability that the representative sells more than 1 policy?
Hi
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re: BINOMIAL Probability Distribution
Note: The probability of x successes in n trials is:
P = nCx* where p and q are the probabilities of success and failure respectively.
In this case p =.8 and q = .2 and n = 4
nCx =
P(0) = (.2)^4 = .0016
P(1) = 4(.8)^1(.2)^3 = 0.0256
P(2) = 6(.8)^2(.2)^2= 0.1536
P(3) = 4(.8)^3(.2)^1 = 0.4096
P(4) = (.8)^4 = 0.4096
Mean = np = 4*.8
Variance = npq = 4*.8*.2
P(at least one) = 1- P(0)
P(more than one) = 1 - P0) - P(1)
