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The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.04
1 answer:
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()
The mean of Poisson distribution is given by, E(X) = = 0.04
which means, X ~ Poisson(0.04)
The probability distribution function of a Poisson random variable is:
Now, we know that 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, 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) = =
= 0.6703
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Since we have it as the largest square, then its largest possible modulus of "x" will be when </span>
<span>, that is, when </span>
<span>.
Since it is the largest possible module, the rightmost number of the "0" in the line of integers, with the largest number being x = 9 and the leftmost number of the "0" in the line of integers, , Will be </span>
<span>
Therefore, the difference requested in the statement is:
</span>
Since we have it as the largest square, then its largest possible modulus of "x" will be when </span>
Since it is the largest possible module, the rightmost number of the "0" in the line of integers, with the largest number being x = 9 and the leftmost number of the "0" in the line of integers, , Will be </span>
Therefore, the difference requested in the statement is:
</span>
Answer:
If we compare the p value obtained and the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .
Step-by-step explanation:
1) Data given and notation
n=180 represent the random sample taken
X=125 represent the number of americans between 17 to 24 that not qualify for the military
estimated proportion of americans between 17 to 24 that not qualify for the military
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
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 less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 provided . The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
If we compare the p value obtained and the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .