Question

3-53. A device that monitors the levels of pollutants has WILEYO GO sensors that detect the amount of CO in the air. Placed in a particular location, it is known that the amount of CO is nor- mally distributed with a mean of 6.23 ppm and a variance of 4.26 ppm2 (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 ppm and 8.5 ppm? (c) An alarm is to be activated if the CO levels exceed a cer- tain threshold. Specify the threshold such that it is 3.75 standard deviations above the mean

Answer 1

Solution :

Given that ,

mean = $\mu$ = 6.23 ppm

variance =  $\sigma$2 = 4.26 ppm

standard deviation = $\sigma$ = $\sqrt{}$$\sigma$2 = $\sqrt{}$4.26 = 2.06 ppm

a) P(x > 9) = 1 - p( x< 9 )

=1- p P[(x - $\mu$ ) / $\sigma$ < (9 - 6.23) / 2.06 ]

=1- P(z < 1.34)

Using z table,

= 1 - 0.9099

= 0.0901

b) P( 5.5 < x < 8.5 ) = P[(5.5 - 6.23)/ 2.06) < (x - $\mu$ ) /$\sigma$  < (8.5 - 6.23) / 2.06) ]

= P( -0.35 < z < 1.10)

= P(z < 1.10) - P(z < -0.35)

Using z table,

= 0.8643 - 0.3632

= 0.5011

c) z = 3.75 ( standard deviation above the mean )

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 3.75 * 2.06 + 6.23

x = 13.96 ppm