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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
, 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.
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:
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
Answer: The percentage of the original price was his discount = 10%
Step-by-step explanation:
Given: original price = $60
Price after discount = $54
Discount = Price after discount - original price
= $ (60-54) = $6
The percentage of the original price was his discount =
Hence, the percentage of the original price was his discount = 10%