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.

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