Data: 40% (0.40) of policyholders 55 years or older submit a claim during the ye
ID: 3149436 • Letter: D
Question
Data: 40% (0.40) of policyholders 55 years or older submit a claim during the year
15 policyholders are randomly selected for company records.
Problem:
(a) How many of the policyholders would you expect to have filed a claim within the last year?
Solution: 40% of the 15 = 6
Answer: 6 policyholders would be expected to have filed a claim within the last year.
Please help me to understand problem (b) below.
Problem:
(b) What is the probability that 10 of the selected policyholders submitted a claim last year?
I have found an answer of P(x=10) = 15 C 10 (0.4)^10 (0.6)^5 = 0.0.45
I understand that the 15 refers to the number of policyholders and the 10 refers to selected policyholders, but I do not understand where they came up with the (0.4)^10 or the (0.6)^5.
The formula in the book is P(x) = nCx x (1 – )n – x, with n = the number of trials; x is the random variable defined as the number of successes; and is the probability of a success on each trial. I do not understand how this formula applies to problem (b) above.
Explanation / Answer
THE POPULATION PROBABILITY to clai, = 0.4
A) THEREFORE THE XPECTED = 0.4*15 = 6.
AS SOLVED BY U AND U NEED SOLUTION FOR BASICALLY PART B
PART B)
HERE THE FORMULA TO BE USED = nCr*(p)^r*(q)^(n-r)
in this part the p = 0.4 as given probabiliy
q = 1-p = 1 -0.4 = 0.6
while n = 15
and x = 10
therefore p(x=10) = 15C10*(0.4)^10*(0.6)^5 = 0.0244
as per ur query the power to 0.4 and the power to 0.6 are 10 and 5 respectively because of the formula of the bernauli.
i state the formula again = p(X=x) = nCx*(p)^x *(q)^(n-x)
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