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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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Jackson buys a grape snow cone on a hot day. By the time he eats all the "snow" off the top, the paper cone is filled with 27\pi
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Question:

Jackson buys a grape snow cone on a hot day. By the time he eats all the "snow" off the top, the paper cone is filled with 27\pi27π27, pi cm^3 3 start superscript, 3, end superscript of melted purple liquid. The radius of the cone is 333 cm. What is the height of the cone?

Answer:

The height of the cone is 9 \ cm

Explanation:

It is given that the radius of the cone is 3 \ cm

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The height of the cone can be determine using the formula,  

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Substituting the values V=27 \pi and r=3, we get,

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2 years ago
The percentage of body fat of a random sample of 36 men aged 20 to 29 found a sample mean of 14.42. Find a 95% confidence interv
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Answer:

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

\bar X=14.42 represent the sample mean

\mu population mean (variable of interest)

\sigma=6.95 represent the population standard deviation

n=36 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that z_{\alpha/2}=1.96

Now we have everything in order to replace into formula (1):

14.42-1.96\frac{6.95}{\sqrt{36}}=12.150    

14.42+ 1.96\frac{6.95}{\sqrt{36}}=16.690    

So on this case the 95% confidence interval would be given by (12.150;16.690)    

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