Question

The number of traffic accidents that occur on a particular stretch of road during a month follows a posson distribution with a mean of 7.6. 1.Find the probability that less than three accidents will occur next month on this stretch of road. 2. Find the probability of observing exactly three accidents. 3. Compare the results of poisson and binomial distribution for the above two probabilities

Answer 1

0.01874  &  0.03658

Step-by-step explanation:

The number of traffic accidents that occur on a particular stretch of road during a month follows a Poisson distribution with a mean of 7.6

λ  =   7.6

P(x)  = λˣ  e^(-λ) / x!

probability that less than three accidents will occur next month on this stretch of road.

= P(0) + P(1) + P(2)

=  7.6⁰  e^(-7.6) / 0!   + 7.6¹  e^(-7.6) / 1!  + 7.6²  e^(-7.6) / 2!

= 5 * 10⁻⁴  + 7.6 * 5 * 10⁻⁴   + 28.88 * 5 * 10⁻⁴

= 5 * 10⁻⁴ ( 1 + 7.6 + 28.88)

= 5 * 10⁻⁴ ( 37.48)

= 187.4* 10⁻⁴

= 0.01874

Find the probability of observing exactly three accidents.

P(3) = 7.6³  e^(-7.6) / 3!

= 73.16 * 5 * 10⁻⁴

= 365.8 * 10⁻⁴

= 0.03658

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Answer 2

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The number of traffic accidents that occur on a particular stretch of road during a month follows a poisson distribution with a mean of 7.6

λ = 7.6

P(x) = λˣ e^(-λ) / x!

probability that less than three accidents will occur next month on this stretch of road.

= P(0) + P(1) + P(2)

= 7.6⁰ e^(-7.6) / 0! + 7.6¹ e^(-7.6) / 1! + 7.6² e^(-7.6) / 2!

= 5 * 10⁻⁴ + 7.6 * 5 * 10⁻⁴ + 28.88 * 5 * 10⁻⁴

= 5 * 10⁻⁴ ( 1 + 7.6 + 28.88)

= 5 * 10⁻⁴ ( 37.48)

= 187.4* 10⁻⁴

= 0.01874

Find the probability of observing exactly three accidents.

P(3) = 7.6³ e^(-7.6) / 3!

= 73.16 * 5 * 10⁻⁴

= 365.8 * 10⁻⁴

= 0.03658