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Papessa [141]
2 years ago
7

Tire pressure monitoring systems (TPMS) warn the driver when the tire pressure of the vehicle is 26% below the target pressure.

Suppose the target tire pressure of a certain car is 31 psi (pounds per square inch.)(a) At what psi will the TPMS trigger a warning for this car?(b) Suppose tire pressure is a normally distributed random variable with a standard deviation equal to 2 psi. If the car's average tire pressure is on target, what is the probability that the TPMS will trigger a warning?
Mathematics
1 answer:
ZanzabumX [31]2 years ago
6 0

Answer:

a) 22.94 psi

b) 5.93\times10^{-5}

Step-by-step explanation:

a)The pressure at which will trigger a warning is

31 - 31*0.26 = 22.94 psi

b) The probability that that the TPMS will trigger warning at 22.94 psi, given that tire pressure has a normal distribution with average of 31 psi and standard deviation of 2 psi

f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}

where x = 22.94, \mu = 31, \sigma = 2

f(22.94)={\frac {1}{2 {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {22.94-31}{2 }}\right)^{2}}

f(22.94)=0.2e^{-8.12} = 5.93\times10^{-5}

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Dahasolnce [82]

Answer:

P(5.5

And we can find this probability using the normal standard distribution or excel and we got:

P(-2.769

Step-by-step explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(5.86,0.13)  

Where \mu=5.86 and \sigma=0.13

We are interested on this probability

P(5.5

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(5.5

And we can find this probability using the normal standard distribution or excel and we got:

P(-2.769

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Let X is the random number Erik thinks of, and Y is the random number Nita thinks of.
Both X and Y are in the range from 0 to 20.
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The difference between the two numbers can be written X-Y, or Y-X depending on which number (X or Y) is greater. But we do not know that. In order not to get negative value, we calculate absolute value of X-Y,  written |X-Y| which will give positive value whether X is greater than Y or not.
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