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 <em>lognormal </em>distribution with mean  and standard deviation
 and standard deviation  , the z-score of a measure X is given by:
, the z-score of a measure X is given by:
 
- 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  . .
- The standard deviation is of  
Question 1:
The 90th percentile is X when Z has a p-value of 0.9, hence <u>X when Z = 1.28.</u>






The 90th percentile of this distribution is of 136 dB.
Question 2:
The probability is the <u>p-value of Z when X = 150</u>, hence:



 has a p-value of 0.9147.
 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
 
 
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  , and we want to find P(X = 6), then:
, and we want to find P(X = 6), then:


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
 
        
             
        
        
        
Y = kx
72 = k*3
73/3 = k
24 = k
y = 24x
        
             
        
        
        
Answer:
k12
Step-by-step explanation: