Question

The hourly median power (in decibels) of received radio signals transmitted between two cities

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{\ln{X} - \mu}{\sigma}$

• 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 $\mu = 3.5$.

• The standard deviation is of $\sigma = \sqrt{1.22}$

Question 1:

The 90th percentile is X when Z has a p-value of 0.9, hence X when Z = 1.28.

$Z = \frac{\ln{X} - \mu}{\sigma}$

$1.28 = \frac{\ln{X} - 3.5}{\sqrt{1.22}}$

$\ln{X} - 3.5 = 1.28\sqrt{1.22}$

$\ln{X} = 1.28\sqrt{1.22} + 3.5$

$e^{\ln{X}} = e^{1.28\sqrt{1.22} + 3.5}$

$X = 136$

The 90th percentile of this distribution is of 136 dB.

Question 2:

The probability is the p-value of Z when X = 150, hence:

$Z = \frac{\ln{X} - \mu}{\sigma}$

$Z = \frac{\ln{150} - 3.5}{\sqrt{1.22}}$

$Z = 1.37$

$Z = 1.37$ 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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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 $p = 0.9147, n = 10$, and we want to find P(X = 6), then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.9147)^{6}.(0.0853)^{4} = 0.0065$

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 brainly.com/question/24863377