Question

# Airliners taking off from City Central Airport produce a mean noise level of 108 dB (decibels),...

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?

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

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