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Talja [164]
3 years ago
6

Suppose that the average number of accidents occurring on a highway each day is 3.5. Find the probability that at least two acci

dents occur today. Find the probability that at most one accident occurs today.
Mathematics
2 answers:
BlackZzzverrR [31]3 years ago
7 0

Answer: 0.1359

Step-by-step explanation:

This is a Poisson distribution. The formula for calculating Poisson distribution is given as :

P (X = x) = \frac{e^{-λ}λ^{x}}{x!}

λ = 3.5

To find the probability that at most one accident occur today implies that , accident might not happen at all , the maximum accident that can happen is 1, substituting this into the formula , we have

P(X=0) = \frac{e^{-3.5}3.5^{0} }{0!}

P(X=0) = 0.0302

P(X=1) =   \frac{e^{-3.5}3.5^{1} }{1!}

P(X=1) = 0.1057

Therefore , the probability that at most one accident occurs today.

= 0.0302 + 0.1057

= 0.1359

P(x=2) =   \frac{e^{-3.5}3.5^{2} }{2!}

P(X=2) = 0.1850

irina1246 [14]3 years ago
3 0

Answer:

48 yards

Step-by-step explanation:

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Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

       P(a < X < b) =  \int_{a}^{b} \frac{2}{x^{3}} dx = 2\int_{a}^{b} x^{-3} dx

                            = 2[ \frac{x^{-3+1} }{-3+1}]^{b}_a   dx    { Because \int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a }

                            = 2[ \frac{x^{-2} }{-2}]^{b}_a = \frac{2}{-2} [ x^{-2} ]^{b}_a

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a) Now P(X < 5) = P(1 < X < 5)  {because x > 1 }

     Comparing with general probability we get,

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b) P(X > 8) = P(8 < X < ∞) = 1/8^{2} - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = \frac{1}{6^{2} } - \frac{1}{10^{2} } = \frac{1}{36} - \frac{1}{100 } = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

                                = (\frac{1}{1^{2} } - \frac{1}{6^{2} }) + (1/10^{2} - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

               ⇒  P(1 < X < x) = 0.75

               ⇒  \frac{1}{1^{2} } - \frac{1}{x^{2} } = 0.75

               ⇒  \frac{1} {x^{2} } = 1 - 0.75 = 0.25

               ⇒  x^{2} = \frac{1}{0.25}   ⇒ x^{2} = 4 ⇒ x = 2  

Therefore, value of x such that P(X < x) = 0.75 is 2.

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Answer:

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Answer:

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Step-by-step explanation:

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