Question

Airliners taking off from City Central Airport produce a mean noise level of 108

dB (decibels), with a standard deviation of 6.7 dB. To encourage airlines to

refit their aircraft with quieter engines, any airliners with a noise level above

120 dB must pay a “nuisance fee”.

a.

What percent of the airliners will be billed a nuisance fee?

b.

After two years, a sample of 1000 airliners showed that 3 were billed

the nuisance fee. Assuming that the program has been effective, and

that the standard deviation has not changed, what is the new mean

noise level?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	108
std deviation   =σ=	6.700

percent of the airliners will be billed a nuisance fee:

probability =P(X>120)=P(Z>(120-108)/6.7)=P(Z>1.79)=1-P(Z<1.79)=1-0.9633=0.0367 ~ 3.67%

b)

since top (3/1000)~ 3% will pay nuisance fee which is at 97th percentile:

for 97th percentile critical value of z=

1.88

therefore new mean =X-z*standard deviation =120-1.88*6.7 =107.404 dB