All I want to know is how they got the standard deviation of 3

We want to find the psi that will trigger a warning, and we know that the warning is triggered when the psi is 28% below the target pressure. The target pressure for the car that we are looking at is 32 psi. So, 28% below this is:

32 - (.28*32) = 32 - 8.96 = 23.04 psi

So the TPMS will trigger a warning for this car when the tire pressure is below 23.04 psi.

(b) We want to find the probability that the warning will be triggered. For this car, that means that we need to find the probability that the psi is 23.04 or lower. We can calculate a z-score by using the following formula:

z=standarddeviationxmean

The mean will be our target pressure of 32 psi, the standard deviation is 3 psi and we will use 23.04 psi for x:

z=323.0432= 38.96= -2.99

A tire pressure of 23.04 psi corresponds to a z-score of -2.99. We need the probability that the psi is 23.04 or lower, which corresponds to finding the area to the left of z = -2.99. We can use a table of areas under the standard normal curve. When we look up our z-score, we find that the

area to the left of z = -2.99 is 0.0014

So the probability that the TPMS will trigger a warning is 0.0014.

We can also find this answer by using technology. In Excel, we can type in the following command:

= NORM.DIST(23.04, 32, 3, TRUE)

This also gives us an answer of 0.0014 (when rounded to four decimal places).

(c) We want to find the probability that the psi is between 30 and 34. We can calculate z-scores for both of these tire pressures:

z=33032= 32= -0.67 (rounded to two decimal places)

z=33432= 32= 0.67 (rounded to two decimal places)

A tire pressure of 30 corresponds to a z-score of -0.67 and a tire pressure of 34 corresponds to a z-score of 0.67. We want to find the probability that the tire pressure is between 30 and 34, which corresponds to finding the area between z = -0.67 and z = 0.67. We can look up both z-scores in our table of areas to find that the:

area to the left of z = -0.67 is 0.2514

area to the left of z = 0.67 is 0.7486

We can subtract the area to the left of z = -0.67 from the area to the left of z = 0.67 to find the area between the z-scores:

0.7486 - 0.2514 = 0.4972

So the probability that the tire's inflation is within the recommended range is 0.4972.

We can find a more accurate answer by using technology. In Excel, we can type in the following command:

= NORM.DIST(34, 32, 3, TRUE) - NORM.DIST(30, 32, 3, TRUE)

This gives us a more accurate answer of 0.4950 (when rounded to four decimal places).