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Tcecarenko [31]
2 years ago
13

In the past, schools in Los Angeles County have closed an average of three days each year for weather emergencies. What is the p

robability that schools in Los Angeles County will close for four days next year
Mathematics
1 answer:
luda_lava [24]2 years ago
8 0

The probability that the schools in Los Angeles will be closed for four days next year is 0.1680.

<h3>What is Poisson's distribution formula?</h3>

In order to comprehend separate events that happen at a steady pace over the course of a specified amount of time, Poisson's distributions are frequently utilized.

Formula for Poisson's distribution:

f(x) = ((λ^x)/x!).(e^(-λ))

x = Event to be calculated for

λ = Expected number of occurrences

P(X = x) = ((λ^x)/x!).(e^(-λ))

Here λ = 3 and x = 4

P(X = 4) = (81 / 4!).(e^(-3))

P(X = 4) = (81 * e^(-3))/24

P(X = 4) = (81 * 0.049)/24

P(X = 4) = 0.1680

Hence, the probability that schools in Los Angeles will be closed for four days next year is 0.1680.

Learn more about probability on:

brainly.com/question/25870256

#SPJ4

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a) P=0.2

b) P=0.42

c) P=0.7

d) Marginal probability distribution of x:

p_x(0)=p(0,0)+p(0,1)+p(0,2)=0.10+0.04+0.02=0.16\\\\p_x(1)=p(1,0)+p(1,1)+p(1,2)=0.08+0.2+0.06=0.34\\\\p_x(2)=p(2,0)+p(2,1)+p(2,2)=0.06+0.14+0.30=0.50

Marginal probability distribution of y:

p_y(0)=p(0,0)+p(1,0)+p(2,0)=0.10+0.08+0.06=0.24\\\\p_y(1)=p(0,1)+p(1,1)+p(2,1)=0.04+0.2+0.14=0.38\\\\p_y(2)=p(0,2)+p(1,2)+p(2,2)=0.02+0.06+0.3=0.38

e) The two variables are dependant.

Step-by-step explanation:

The joint pmf of X and Y appears in the accompanying tabulation (attached).

a) Is the proability that one hose being used in the self service island and one hose being use in the full service island.

P(X=1 \& Y=1)=P(1,1)=0.2

(according to the tabulation)

b)

P(X\leq1 \& Y\leq1)=P(0,0)+P(1,0)+P(0,1)+P(1,1)\\\\P(X\leq1 \& Y\leq1)=0.10+0.08+0.04+0.20\\\\P(X\leq1 \& Y\leq1)=0.42

c) This event means that at least one hose is in use in each of the island.

P(X\neq0 \& Y\neq0)=P(1,1)+P(1,2)+P(2,1)+P(2,2)\\\\P(X\neq0 \& Y\neq0)=0.20+0.06+0.14+0.30=0.70

d)

Marginal probability distribution of x:

p_x(0)=p(0,0)+p(0,1)+p(0,2)=0.10+0.04+0.02=0.16\\\\p_x(1)=p(1,0)+p(1,1)+p(1,2)=0.08+0.2+0.06=0.34\\\\p_x(2)=p(2,0)+p(2,1)+p(2,2)=0.06+0.14+0.30=0.50

Marginal probability distribution of y:

p_y(0)=p(0,0)+p(1,0)+p(2,0)=0.10+0.08+0.06=0.24\\\\p_y(1)=p(0,1)+p(1,1)+p(2,1)=0.04+0.2+0.14=0.38\\\\p_y(2)=p(0,2)+p(1,2)+p(2,2)=0.02+0.06+0.3=0.38

Compute P(X≤1)

P(X\leq1)=p_x(0)+p_x(1)=0.16+0.34=0.50

e) For X and Y to be independent variables, the following conditions must be true:

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We evaluate the second probability for X=1 and Y=1.

P(x=1 \cap  y=1) = 0.2\\\\p_x(1)*p_y(1)=0.34*0.38=0.1292\\\\\\P(x=1 \cap  y=1)\neq p_x(1)*p_y(1)

This condition is not satisfied, so the two variables are dependant.

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