A device that monitors the levels of pollutants has sensors that detect the amou

Question

A device that monitors the levels of pollutants has sensors that detect the amount of CO in the air. Placed in a particular location, it is known that the amount of CO is normally distributed with a mean of 6.23 ppm and a variance of 4.26 ppm2
(a) What is the probability that the CO levels exceed 9.0 ppm? Round your answers to four decimal place (e.g. 98.7654).
(b) What is the probability that the CO level is between 1.5 ppm and 12.5 ppm? Round your answers to four decimal place (e.g. 98.7654).
(c) An alarm is to be activated if the CO levels exceed a certain threshold. Specify the threshold such that it is 2.75 standard deviations above the mean.
If needed, round z-score to 2 decimal places and use Appendix A, Table I.

Explanation / Answer

Mean = 6.23

Variance = 4.26

SD = sqrt(4.26) = 2.06

z=(X-Mean)/SD

a)
z = (9-6.23)/2.06 = 1.34
The area under the standard normal curve right to the z value indicates the required probability.
Required probability = P(X>9)
=P(z>1.34)
= 0.0901 Answer

b)
z1 = (1.5-6.23)/2.06 = -2.3
z2 = (12.5-6.23)/2.06 = 3.04
The area under the standard normal curve between these two z values indicates the required probability.
Required probability = P(1.5<X<12.5)
=P(-2.3<z<3.04)
=P(z<3.04) - P(z<-2.3)
=0.9988 - 0.0107
= 0.9881 Answer

c)
Required Threshold = Mean + 2.75*SD
=6.23+2.75*2.06
=11.895 Answer

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