Confused how to do this probability question

Question

The hourly median power (in decibels) of received radio signals transmitted between two cities follows a lognormal distribution with parameter values u = 3.5 and o(sigma)=1.2

a. What is the 90th percentile of this distribution?

b. What is the probability that received power for one of these radio signals is

less than 150 decibels?

c. Consider a random sample of 10 radio signals. What is the probability that for 6 of these signals, the received power is less than 150 decibels?

Answer 1

a) We need the Z-score for 0.9 (90%), which is 1.2816 (Normal Distribution).

We need to solve for X: (X-μ)/σ=1.2816 using the given μ, σ:

X=μ+1.2816σ=3.5+1.2816×1.2=5.038 approx.

But because this is a lognormal distribution, we need e^5.038=154.15. Round this to 154dB as the 90^th percentile.

b) ln(150)=5.011 approx. Now we use the Normal Distribution and this value for X.

Z=(5.011-3.5)/1.2=1.2589 approx. corresponding to 0.896 or 89.6%. So the probability that the received power is less than 150dB is 89.6%.

c) The probability of received power being less than 150dB is 89.6% (see b)). In a random sample of 10 the number of ways of selecting 6 is 210. There is a 10.4% probability of received power >150dB. For exactly 6 out of 10 being <150dB = 210(0.896⁶)(0.104⁴)=0.0127 approx=1.27%.