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Of all cities in the United States, Amherst, New York, has the fewest number of days...

Of all cities in the United States, Amherst, New York, has the fewest number of days per year clear of clouds, 4.4. Other cities with very few clear days include Buffalo, New York, Lakewood, Washington, and Seattle, Washington. Suppose a random year is selected. (Round your answers to four decimal places.)

(a)

What is the probability that Amherst will have exactly four days clear of clouds?

(b)

What is the probability of fewer than six days clear of clouds?

(c)

What is the probability of at least nine days clear of clouds?

(d)

Suppose that between 2 and 11 (inclusive) days are clear of clouds. What is the probability of more than five days clear of clouds?

this is all the data provided for this question.

Solutions

Expert Solution

Answer:

a)

Given,

= 4.4

Here we are utilizing poisson distribution

P(X = 4) = e^-4.4 * 4.4^4 / 4!

= 4.6888/24

P(X = 4) = 0.19174

Hence the probability that Amherst will have exactly four days clear of clouds is 0.19174

b)

To determine the probability of fewer than six days clear of clouds

P(X < 6) = e^-4.4*4.4^x / x!

= e^-4.4*4.4^0/0! + e^-4.4*4.4^1/1! + e^-4.4*4.4^2/2! + e^-4.4*4.4^3/3! + e^-4.4*4.4^4/4! + e^-4.4*4.4^5/5!

= 0.012277339 + 0.054020295 + 0.11884465 + 0.174305487 + 0.191736035 + 0.168727711

P(X < 6) = 0.7199

c)

To determine the probability of at least nine days clear of clouds

P(X >= 9) = 1 - P(X <= 8)

= 1 - [e^-4.4*4.4^0/0! + e^-4.4*4.4^1/1! + e^-4.4*4.4^2/2! + e^-4.4*4.4^3/3! + e^-4.4*4.4^4/4! + e^-4.4*4.4^5/5! + e^-4.4*4.4^6/6! + e^-4.4*4.4^7/7! + e^-4.4*4.4^8/8!]

= 1 - [ 0.012277339 + 0.054020295 + 0.11884465 + 0.174305487 + 0.191736035 + 0.168727711 + 0.123733655 + 0.07777544 + 0.042776492]

= 1 - [0.9642]

P(X >= 9) = 0.0358

d)

To give required probability

P(2 <= X <= 11) = e^-4.4 * 4.4^x / x!

On solving we get

= 0.9317

P(2 <= X <= 11) = 0.9317

milcah answered 10 months ago
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