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Sagot :
Using the lognormal and the binomial distributions, it is found that:
- The 90th percentile of this distribution is of 136 dB.
- There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.
- There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.
In a lognormal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{\ln{X} - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
In this problem:
- The mean is of [tex]\mu = 3.5[/tex].
- The standard deviation is of [tex]\sigma = \sqrt{1.22}[/tex]
Question 1:
The 90th percentile is X when Z has a p-value of 0.9, hence X when Z = 1.28.
[tex]Z = \frac{\ln{X} - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{\ln{X} - 3.5}{\sqrt{1.22}}[/tex]
[tex]\ln{X} - 3.5 = 1.28\sqrt{1.22}[/tex]
[tex]\ln{X} = 1.28\sqrt{1.22} + 3.5[/tex]
[tex]e^{\ln{X}} = e^{1.28\sqrt{1.22} + 3.5}[/tex]
[tex]X = 136[/tex]
The 90th percentile of this distribution is of 136 dB.
Question 2:
The probability is the p-value of Z when X = 150, hence:
[tex]Z = \frac{\ln{X} - \mu}{\sigma}[/tex]
[tex]Z = \frac{\ln{150} - 3.5}{\sqrt{1.22}}[/tex]
[tex]Z = 1.37[/tex]
[tex]Z = 1.37[/tex] has a p-value of 0.9147.
There is a 0.9147 = 91.47% probability that received power for one of these radio signals is less than 150 decibels.
Question 3:
10 signals, hence, the binomial distribution is used.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
For this problem, we have that [tex]p = 0.9147, n = 10[/tex], and we want to find P(X = 6), then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{10,6}.(0.9147)^{6}.(0.0853)^{4} = 0.0065[/tex]
There is a 0.0065 = 0.65% probability that for 6 of these signals, the received power is less than 150 decibels.
You can learn more about the binomial distribution at https://brainly.com/question/24863377
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