Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
