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alexandr1967 [171]
2 years ago
10

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.04

flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel. (a) What is the probability that there are no surface flaws in an auto's interior
Mathematics
1 answer:
Furkat [3]2 years ago
4 0

Answer:

(a) Probability that there are no surface flaws in an auto's interior is 0.6703 .

Step-by-step explanation:

We are given that the number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.04 flaws per square foot of plastic panel.

Let X = Distribution of number of surface flaws in plastic panels

So, X ~ Poisson(\lambda)

The mean of Poisson distribution is given by, E(X) = \lambda = 0.04

which means, X ~ Poisson(0.04)

The probability distribution function of a Poisson random variable is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!}; for  x=0,1,2,3...

Now, we know that \lambda for per square foot of plastic panel is 0.04 and we are given that an automobile interior contains 10 square feet of plastic panel.

Therefore, \lambda for 10 square foot of plastic panel is = 10 * 0.04 = 0.4

(a) Probability that there are no surface flaws in an auto's interior =P(X=0)

     P(X = 0) = \frac{e^{-0.4}*0.4^{0}}{0!} = e^{-0.4} = 0.6703

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99/9081 =                               => as 9081 = 9 * 1009 = 3 * 3 * 1009

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z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=18 represent the ounce cups of coffee that were underfilled

\hat p=\frac{18}{100}=0.18 estimated proportion of ounce cups of coffee that were underfilled

p_o=0.1 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion it's higher than 0.1 or 10%:  

Null hypothesis:p\leq 0.1  

Alternative hypothesis:p > 0.1  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed for this case is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a one right tailed test the p value would be:  

p_v =P(Z>2.67)=0.0037  

So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance thetrue proportion is not significanlty higher than 0.1 or 10% .  

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