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 9.5 ppm? Round your answers to four decimal place (e.g. 98.7654).
(b) What is the probability that the CO level is between 5.5 ppm and 8.5 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

the distribution is normal

the mean = 6.23

variance = 4.26

standard deviation = sqrt(variance) = sqrt(4.26) = 2.06

the formula will be used = z = (x-mean)/standard deviation

A) p(x>9.5) =

For x = 9.5, z = (9.5 - 6.23) / 2.06 = 1.58

Hence P(x > 9.5) = P(z > 1.58) = [total area] - [area to the left of 1.58]

1 - [area to the left of 1.58]

now from the z table we will take the value of z score = 1.58

    = 1 - 0.9429 = 0.0571

b) P(5.5<x<8.5) =

For x = 5.5 , z = (5.5 - 6.23) / 2.06 = -0.35 and for x = 8.5, z = (8.5 - 6.23) / 2.06 = 1.10

Hence P(5.5 < x < 8.5) = P(-0.35 < z < 1.10) = [area to the left of z = 1.10] - [area to the left of -0.35]

= 0.8643 - 0.3632 = 0.5011

c) here x = 2.75

formula to be used : x = mean + z*standard deviation

therefore

x = 6.23+2.75*2.06

therefore x = 11.895

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