Question

In: Statistics and Probability

The random variable X follows a Poisson process with the given mean. Assuming mu equals 6...

The random variable X follows a Poisson process with the given mean. Assuming

mu equals 6 commaμ=6,

compute the following.

​(a)​ P(66​)

​(b)​ P(Xless than<66​)

​(c) ​P(Xgreater than or equals≥66​)

​(d) ​P(55less than or equals≤Xless than or equals≤77​)

Solutions

Expert Solution

SOLUTION:

From given data,

The random variable X follows a Poisson process with the given mean. Assuming mu equals 6 comma compute the following. Please show work.

(a) P(6)

(b) P(X less than 6)

(c) P(X greater than or equals 6)

(d) P(5 less than or equals X less than or equals 7)

X Poisson ( )

P(X) = e- * x / x !

= 6

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

(a) P(6)

x = 6

P(6) = e-6 * 66 / 6 !

P(6) = 0.0024787* 46656/ 720

P(6) = 115.6486615 / 720

P(6) = 0.160623

(b) P(X less than 6)

P(X < 6) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)

P(X < 6) = e-6 * 60 / 0 !+e-6 * 61 /1 !+e-6 * 62 / 2 !+e-6 * 63 / 3 !+e-6 * 64 / 4 !+e-6 * 65 / 5 !

P(X < 6) = e-6+ e-6 * 6 /1 +e-6 * 36 / 2+e-6 * 216 / 6+e-6 * 1296 / 24+e-6 * 7776/ 120

P(X < 6) = 0.0024787+ 0.0148722+0.0446166+0.0892332+0.1338498+0.16061976

P(X < 6) = 0.445670

(c) P(X greater than or equals 6)

P(X > 6 ) = 1 - P(X < 6)

P(X > 6 ) = 1 - 0.445670

P(X > 6 ) = 0.55433

(d) P(5 less than or equals X less than or equals 7)

P(5 < X < 7) = P(5)+P(6)+P(7)

P(5 < X < 7) = e-6 * 65 / 5 !+e-6 * 66 / 6 !+e-6 * 67 / 7!

P(5 < X < 7) = e-6 * 7776/ 120+e-6 * 46656 / 720+e-6 * 279936 / 5040

P(5 < X < 7) = 0.16061976+0.16061976+0.13767408

P(5 < X < 7) = 0.458913


Related Solutions

Suppose x is a normally distributed random variable with mu equals 43 and sigma equals 5....
Suppose x is a normally distributed random variable with mu equals 43 and sigma equals 5. Find a value x 0 of the random variable x that satisfies the following equations or statements. a. ​P(x less than or equals x 0​)equals0.8413 b. ​P(x greater thanx 0​)equals0.025 c. ​P(x greater thanx 0​) equals 0.95 d.​ P(28 less than or equals x less thanx 0​) equals 0.8630 e.​ 10% of the values of x are less than x 0. f.​ 1% of...
Assuming a random variable X is distributed with a mean of the second digit of your...
Assuming a random variable X is distributed with a mean of the second digit of your student number, a variance of the last digit of your student number divided by 10 ( if the last digit is zero, use 9). If you take 64 independent samples from X, what is the probability that the sample means is greater than 5? The sample mean that you get from the previous sample is 3.5; please construct a 95% confidence interval based on...
Let Y denote a random variable that has a Poisson distribution with mean λ = 6....
Let Y denote a random variable that has a Poisson distribution with mean λ = 6. (Round your answers to three decimal places.) (a) Find P(Y = 9). (b) Find P(Y ≥ 9). (c) Find P(Y < 9). (d) Find P(Y ≥ 9|Y ≥ 6).
1. If the random variable x has a Poisson Distribution with mean μ = 53.4, find...
1. If the random variable x has a Poisson Distribution with mean μ = 53.4, find the maximum usual value for x. Round your answer to two decimal places. 2. In one town, the number of burglaries in a week has a Poisson distribution with mean μ = 7.2. Let variable x denote the number of burglaries in this town in a randomly selected month. Find the smallest usual value for x. Round your answer to three decimal places. (HINT:...
The random variable X follows a normal distribution with a mean of 10 and a standard...
The random variable X follows a normal distribution with a mean of 10 and a standard deviation of 3. 1. What is P(7≤X≤13)? Include 4 decimal places in your answer. 2. What is the value of k such that P(X>k)=0.43? Include 2 decimal places in your answer.
Assume that? women's heights are normally distributed with a mean given by mu equals 64.6 in?,...
Assume that? women's heights are normally distributed with a mean given by mu equals 64.6 in?, and a standard deviation given by sigma equals 1.9 in. ?(a) If 1 woman is randomly? selected, find the probability that her height is less than 65 in. ?(b) If 50 women are randomly? selected, find the probability that they have a mean height less than 65 in. ?(?a) The probability is approximately nothing. ?(Round to four decimal places as? needed.) ?(b) The probability...
Assume that​ women's heights are normally distributed with a mean given by mu equals 63.4 in​,...
Assume that​ women's heights are normally distributed with a mean given by mu equals 63.4 in​, and a standard deviation given by sigma equals 1.9 in. ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 64 in. ​(b) If 31 women are randomly​ selected, find the probability that they have a mean height less than 64 in.
Assume that​ women's heights are normally distributed with a mean given by mu equals 64.4 in...
Assume that​ women's heights are normally distributed with a mean given by mu equals 64.4 in and a standard deviation given by sigma equals 2.9 in (a) If 1 woman is randomly​ selected, find the probability that her height is less than 65 in. ​(b) If 45 women are randomly​ selected, find the probability that they have a mean height less than 65
Assume that​ women's heights are normally distributed with a mean given by mu equals 62.1 in​,...
Assume that​ women's heights are normally distributed with a mean given by mu equals 62.1 in​, and a standard deviation given by sigma equals 2.7 in. ​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in. ​(b) If 32 women are randomly​ selected, find the probability that they have a mean height less than 63 in.
A random sample is selected from a normal population with a mean mu equals 30 space...
A random sample is selected from a normal population with a mean mu equals 30 space a n d space s tan d a r d space d e v i a t i o n space sigma equals 8. After a treatment is administered to the individuals in the sample, the sample mean is found to be M=33. (a) If the sample consists of n=16 scores is the sample mean sufficient to conclude that the treatment has a significant...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT