Question #253054

1)                 15 policyholders over 50 are selected at random; what is the probability that at least 10 will submit a claim during the year? What is the probability that 4 will submit a claim during the year? How many do you expect to submit a claim. What is the standard deviation?



1
Expert's answer
2021-10-19T09:44:13-0400

The number of policyholders that submit a claim can be described as binomial distribution X = Bin(15, 0.3)

The probability that at least 10 will submit a claim during the year:

P(X10)=P(X=10)+...+P(X=15)=0.003+0.0006+0.0001+0+0+0=P(X≥10) = P(X=10)+...+P(X=15) = 0.003+0.0006+0.0001+0+0+0=

=0.0037= 0.0037 (The last three probabilities is very small and rounded to 4 digits are equal to 0)

The probability that 4 will submit a claim during the year:

P(X=4)=0.2186P(X=4)=0.2186

How many do you expect to submit a claim:

The expected value of Bin(15, 0.3) is the answer

E(Bin(15,0.3)=150.3=4.5E(Bin(15, 0.3) = 15*0.3 = 4.5

If round, it can be assumed that 4 is the most expected value(P(X=4) is a bit more than P(X=5))

What is the standard deviation:

D(Bin(15,0.3)=150.30.7=3.15D(Bin(15, 0.3)=15*0.3*0.7=3.15

Standard deviation is equal to D=3.15=1.77\sqrt{D} =\sqrt{3.15}=1.77


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