Answer: The probability is 0.0005%
Step-by-step explanation:
We are told that the data follows a lognormal distribution, hence we can use the formula for the probability given by:
[tex]f_{X}(x)=\frac{1}{x} \cdot \frac{1}{\sigma \sqrt{2 \pi}} \exp \left(-\frac{(\ln x-\mu)^{2}}{2 \sigma^{2}}\right)[/tex]
Where, in this case, we have [tex]\mu =4 and \sigma=2. To fin the probability we only need to evaluate the formula for x = 270 dB:
[tex]f_{X}(270)=\frac{1}{270} \cdot \frac{1}{2 \times \sqrt{2 \pi}} \exp \left(-\frac{(\ln 270-4)^{2}}{2\times 2^{2}}\right)\\f_{X}(270)= 1.5\times10^{-5} = 0.00005 = 0.0005\%[/tex]
We fin out that the probability that the company uses more thant 270dB during any particular hour is very low, 0.0005%, for the given the parameters μ = 4 and σ = 2 in the lognormal distribution.
