Question #e49ee

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Question #e49ee

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Answer 1 · 2015-09-17T04:39:35

p = 0.3, n= 12, let x = The number of accidents to 18 year old drivers within one year of taking license;
Required Prob. = 0.49071.

Explanation:

X follows binomial distribution with n= 12 and p = 0.3
general p.d.f. is f ( x) = $n C_{x}$ $n C _x$ $p^{x}$ $p^x$ $q^{n - x}$ $q^ (n-x)$ , x = 0,1,2...n and q = 1 - p.
Required Pr{ X $\leq$ $<=$ 3} = Pr { x = 0, 1, 2, 3}
= Pr {X = 0} + Pr{X = 1} + Pr { X = 2} + Pr {X = 3} = $12 C_{0}$ $12 C _0$ ${(0.7)}^{12}$ $(0.7)^12$ + $12 C_{1}$ $12 C _1$ 0.3 ${(0.7)}^{11}$ $(0.7)^11$ + $12 C_{2}$ $12C_2$ ${(0.3)}^{2}$ $(0.3) ^2$ ${(0.7)}^{10}$ $(0.7)^10$ + $12 C_{3}$ $12 C _3$ ${(0.3)}^{3}$ $(0.3)^3$ ${(0.7)}^{9}$ $(0.7)^9$
= 0.01384 + 0.07118 + 0. 16779 + 0.23790 =0.49071

Answer 2 · 2015-09-17T04:55:04

Refer explanation section

Explanation:

Following Binomial Distribution the answer is $P_{x < 3} = 0.4919$ $P_(x<3)=0.4919$
Following Poisson Distribution the answer is $P_{x < 3} = 0.51517$ $P_(x<3)=0.51517$
Following Normal Distribution the answer is $P_{x < 3} = 0.352$ $P_(x<3)=0.352$

I am sorry I have wrongly totaled the probability values.
I have redone the problem using the three distributions.
I have enclosed my work in a pdf file form.

I am unable to get the answer that you have given 0.3327.

May I know in which area the problem appears in your book.

This is the link to the pdf File

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