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 e5.038=154.15. Round this to 154dB as the 90th 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%.