Well it depends. What job is it that your trying to enter?
(This is not the answe)
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:
$300
Step-by-step explanation:
1500 • 20%
1500 • 20/100
15 • 20
300
Hope this helps, good luck! :)
Answer:
Ball Bearing Volume = (4 / 3) * PI * radius ^3
Volume = (4 / 3) * 3.14159265 * (.35)^3
Volume = (4 / 3) * 3.14159265 * 0.042875
Volume = 0.17959438 cc
1 cc of steel = 7.8 g
Volume of 1 kg of steel = 1,000 cc / 7.8 g per cc
1 kg = 128.205 cc
Number of ball bearings we can make = 128.205 / 0.17959438
Equals 713.86
Step-by-step explanation:
Answer:
the answer is r (‑1)/(6*x)-2.1
Step-by-step explanation: