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. Places in a particular location, it is know that the amount of CO is normally distributed with a mean of 6.23 ppm and a variance of 4.26 ppm^2. (a) What is the probability that the CO level exceeds 9 ppm? (b) What is the probability that the CO level is between 5.5 and 8.5 ppm? (c) An alarm is to be activated if the CO level exceed a certain threshold. Specify the threshold such that it is 3.75 standard deviations above the mean.

Explanation / Answer

IT IS GIVEN THAT IT IS NORMALLY DISTRIBUTED

NOW MEAN = 6.23

VARIANCE = 4.26

STANDARD DEVIATION = 4.26^(1/2) = 2.06

THE FORMULA TO BE USED = (X-MEAN)/STANDARD DEVIATION

A) P(X>9)=

For x = 9, z = (9 - 6.23 ) /2.06 = 1.34

Hence P(x > 9) = P(z > 1.34) = [total area] - [area to the left of 1.34]

1 - [area to the left of 1.34]

now from the z table we will take the value of z score = 1.34

    = 1 - 0.9099 = 0.0901

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 THE Z SCORE = 3.75

FORMULA TO BE USED

X = MEAN + Z*STANDARD DEVIATION

X = 6.23 + 3.75*2.06 = 13.955

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