问题及解答

Let $X$= the hourly median power (in decibels (分贝/décibels/Dezibel)) 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. Research National Bureau of Standards, vol. 67D, 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)

\[
E(X)=e^{\mu+\frac{\sigma^2}{2}}=e^{3.5+\frac{1.2^2}{2}}=e^{4.22}\approx 68.03348429
\]

\[
\begin{split}
V(X)&=e^{2\mu+\sigma^2}\cdot(e^{\sigma^2}-1)\\
&=e^{2\times 3.5+1.2^2}\cdot(e^{1.2^2}-1)\\
&=e^{8.44}\cdot(e^{1.44}-1)\\
&\approx 4628.55498456\times(4.22069582-1)\\
&=14907.1676914125565392
\end{split}
\]

Hence, the standard derivation is 

\[
\sqrt{V(X)}=\sqrt{14907.1676914125565392}\approx  122.09491264
\]

(b)

The probability that received power is between $50$ and $250$ dB is

\[
\begin{split}
P(50\leqslant X\leqslant 250)&=P(X\leqslant 250)-P(X\leqslant 50)\\
&=F(250;\mu,\sigma)-F(50;\mu,\sigma)\\
&=\Phi(\frac{\ln(250)-\mu}{\sigma})-\Phi(\frac{\ln(50)-\mu}{\sigma})\\
&=\Phi(\frac{\ln(250)-3.5}{1.2})-\Phi(\frac{\ln(50)-3.5}{1.2})\\
&=\Phi(\frac{5.52146020-3.5}{1.2})-\Phi(\frac{3.91202288-3.5}{1.2})\\
&=\Phi(1.68455017)-\Phi(0.3433524)\\
&\approx\Phi(1.68)-\Phi(0.34)\\
&=0.9535-0.6331\\
&=0.3204
\end{split}
\]

The penultimate equal sign is using value of the Table A.3. in Appendix of the book.

(c)

The probability that $X$ is less than its mean value is

\[
\begin{split}
P(X\leqslant E(X))&=P(X\leqslant 68.03348429)\\
&=\Phi(\frac{\ln(68.03348429)-\mu}{\sigma})\\
&=\Phi(\frac{4.21950771-3.5}{1.2})\\
&=\Phi(0.59958976)\\
&\approx\Phi(0.6)\\
&=0.7257
\end{split}
\]

The skewness implies mean is not equal to median.