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Let $X=$ the hourly median power (in decibels) of received radio signals transmitted between two cities. The authors of the article "Families of Distributions for Hourly Median Power and Instantaneous Power of Received Radio Signals" (J. Res. Nat. Bureau Standards, vol. 67 $\mathrm{D},$ $1963 : 753-762$ argue that the lognormal distribution provides a reasonable probability model for $X .$ If the parameter values are $\mu=3.5$ and $\sigma=1.2,$ calculate the following:(a) The mean value and standard deviation of received power.(b) The probability that received power is between 50 and 250 dB.(c) The probability that $X$ is less than its mean value. Why is this probability not. 5$?$

(a) 68.03 dB, 122.09 dB (b) .3204 (c) .7642, because the lognormal distribution is not symmetric

Intro Stats / AP Statistics

Chapter 3

Continuous Random Variables and Probability Distributions

Section 9

Supplementary Exercises

Continuous Random Variables

Temple University

Cairn University

Idaho State University

Lectures

02:35

Consider the following bin…

01:05

Suppose that $X$ is a norm…

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were given a log. Normal distribution with parameters mu equals 3.5 and Sigma equals 1.2 for a were asked to find the mean and standard deviation of the distribution. Now, for a log normal distribution, the mean is given by the following formula. So if we plug in our values, our perimeter values, we have e t 3.5 plus 1.2 squared, divided by two and this comes out to 68.3 The variance is given by this formula and again plugging in our parameter values. We get a variance of 14,000 907.17 which gives us a standard deviation of 122.9 So that answer is part A. Now for part B were asked for the probability that X is between 50 and 250. So here we have we've taken the log of X and then standardized it to the standard normal distribution because we know it's a log normal. Therefore, the log of X is a normally distributed random variable 0.954 minus 0.634 and this comes out to 0.3 to 0. The probability of 0.3 to 0. That X is between 50 and 2 50 and next. For part C, we're asked for the probability that X is less than its main value. Its mean value from Part A is 68.3 and this comes out to 0.726 This is slightly different than the answer in the back of the textbook. There is an error in the back of the textbook. It seems as though they forgot to divide by 1.2. You here now, the reason why the probability that X is less than the mean isn't exactly 50% is because the log normal distribution is right skewed.

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