Respuesta :
Answer:
The probability that a given user is transmitting is 0.1 = 10%.
The probability that at any given time, exactly n users are transmitting simultaneously is [tex]P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}[/tex].
0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
Each user transmits only 10 percent of the time.
This means that [tex]p = 0.1[/tex]
Find the probability that a given user is transmitting.
The probability that a given user is transmitting is 0.1 = 10%.
Suppose there are 120 users.
This means that [tex]n = 120[/tex]
Find the probability that at any given time, exactly n users are transmitting simultaneously.
This is [tex]P(X = n)[/tex].
This n is different from the n of total number of users(120 in this case) from the standard binomial formula. This is the number of successes, which is the equivalent of x. So
[tex]P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}[/tex]
The probability that at any given time, exactly n users are transmitting simultaneously is [tex]P(X = n) = C_{120,n}.(0.1)^{n}.(0.9)^{120-n}[/tex]
Find the probability that there are 21 or more users transmitting simultaneously.
Now we use the binomial approximation to the normal. We have that:
[tex]\mu = E(X) = np = 120*0.1 = 12[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.1*0.9} = 3.2863[/tex]
Using continuity correction, this probability is [tex]P(X \geq 21 - 0.5) = P(X \geq 20.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 20.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.5 - 12}{3.2863}[/tex]
[tex]Z = 2.59[/tex]
[tex]Z = 2.59[/tex] has a pvalue of 0.9952
1 - 0.9952 = 0.0048
0.0048 = 0.48% probability that there are 21 or more users transmitting simultaneously.
