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BARSIC [14]
2 years ago
5

Suppose the number of hits a webpage receives follows a Poisson distribution. The average number of hits per minute is 2.4. What

is the probability the page will get at least one hit during any given minute?
Mathematics
1 answer:
REY [17]2 years ago
6 0

Answer:

The probability that the page will get at least one hit during any given minute is 0.9093.

Step-by-step explanation:

Let <em>X</em> = number of hits a web page receives per minute.

The random variable <em>X</em> follows a Poisson distribution with parameter,

<em>λ</em> = 2.4.

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

Compute the probability that the page will get at least one hit during any given minute as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-\frac{e^{-2.4}(2.4)^{0}}{0!}\\=1-\frac{0.09072\times1}{1} \\=1-0.09072\\=0.90928\\\approx0.9093

Thus, the probability that the page will get at least one hit during any given minute is 0.9093.

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3 years ago
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

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