Let X = hourly median power (in decibels) of received radio signals transmitted between two cities. It is believed that the lognormal distribution provides a reasonable probability model for X. If the parameter values are $\mu = 3.5 \text { and } \sigma = 1.2 \text {, }$ 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?

Question

Let X = hourly median power (in decibels) of received radio signals transmitted between two cities. It is believed that the lognormal distribution provides a reasonable probability model for X. If the parameter values are $\mu = 3.5 \text { and } \sigma = 1.2 \text {, }$ 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?

Quizplus · Accepted Answer

The Answer is a.

E ( X ) = e ^ { 35 ( 12 ) ^ { 2 } / 2 } = 68.0335 ; V ( X ) = e ^ { 2 ( 3 5 ) + ( 2.1 ) ^ { 2 } } \cdot \left( e ^ { 1.2 ) ^ { 2 } } - 1 \right) = 14907.168

and

\sigma _ { x } = 122.0949

b.

P ( 50 \leq X \leq 250 ) = P \left( z \leq \frac { \ln ( 250 ) - 3.5 } { 1.2 } \right) - P \left( z \leq \frac { \ln ( 50 ) - 3.5 } { 1.2 } \right)

P ( Z \leq 1.68 ) - P ( Z \leq .34 ) = .9535 - .6331 = .3204

c.

P ( X \leq 68.0335 ) = P \left( z \leq \frac { \ln ( 68.0335 ) - 3.5 } { 1.2 } \right) = P ( Z \leq .60 ) = .7257

The lognormal distribution is not a symmetric distribution.

Let X = Hourly Median Power (In Decibels) of Received μ=3.5 and σ=1.2, \mu = 3.5 \text { and } \sigma = 1.2 \text {, }μ=3.5 and σ=1.2,

Let X = Hourly Median Power (In Decibels) of Received $\mu = 3.5 \text { and } \sigma = 1.2 \text {, }$