The hourly median power (in decibels) of received radio signals transmitted between two cities follows a lognormal distribution with parameter values u = 3.5 an
Question
follows a lognormal distribution with parameter values u = 3.5 and o2 = 1.22.
(5 points) What is the 90th percentile of this distribution?
(5 points) What is the probability that received power for one of these radio signals is
less than 150 decibels?
(5 point) Consider a random sample of 10 radio signals. What is the probability that for
6 of these signals, the received power is less than 150 decibels?
1 Answer

1. User Answers joaobezerra
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 zscore 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 zscore, we look at the zscore table and find the pvalue associated with this zscore, 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 pvalue 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 pvalue 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 pvalue 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}.(1p)^{nx}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(nx)!}[/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}.(1p)^{nx}[/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