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 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 ppm and 8.5 ppm? c. An alarm is to be activated if the CO levels exceed a certain threshold. Specify the threshold such that it is 3.25 standard deviations above the mean.

a) 0.09012

b) 0.50116

c) The threshold such that it is 3.25 standard deviations above the mean is 12.41ppm.

And 0.99865 of the CO data range will fall within the range of the threshold.

Only 0.00135 will exceed the threshold.

Step-by-step explanation:

Mean, xbar = 6.23 ppm

Standard deviation = √variance = √4.26 = 2.06ppm

a) To find the probability that the CO level exceeds 9 ppm

We need to standardize 9ppm

z = (x - xbar)/σ = (9 - 6.23)/2.06 = 1.34

P(x > 9) = P(x > 1.34)

We'll use data from the normal probability table for these probabilities

P(x > 9) = P(z > 1.34) = 1 - P(z ≤ 1.34) = 1 - 0.90988 = 0.09012

b) To find the probability that the CO level is between 5.5 ppm and 8.5 ppm

We need to standardize 5.5ppm and 8.5ppm

z = (x - xbar)/σ = (5.5 - 6.23)/2.06 = - 0.35

z = (x - xbar)/σ = (8.5 - 6.23)/2.06 = 1.10

P(5.5 < x < 8.5) = P(-0.35 < z < 1.10)

We'll use data from the normal probability table for these probabilities

P(5.5 < x < 8.5) = P(-0.35 < z < 1.10) = P(z ≤ 1.10) - P(z ≤ -0.35) = 0.86433 - 0.36317 = 0.50116

c) An alarm is to be activated if the CO levels exceed a certain threshold. Specify the threshold such that it is 3.25 standard deviations above the mean.

x = 3σ + xbar = 3(2.06) + 6.23 = 12.41ppm

The amount of CO within the range of the threshold

P(z < 3) = 1 - P(z ≥ 3) = 1 - P(z ≤ -3) = 1 - 0.00135 = 0.99865

Hope this Helps!!!

all of the months have 28 days

hope this : )

-yamauchi wang

55? i think maybe

step-by-step explanation: