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denis23 [38]
3 years ago
8

The lengths of a lawn mower part are approximately normally distributed with a given mean = 4 in. and standard deviation = 0.2 i

n. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.? 34% 68% 95% 99.7%
Mathematics
2 answers:
vova2212 [387]3 years ago
8 0

Answer: EDGE 2020 Quiz

Step-by-step explanation:

1. The average miles per gallon of a particular automobile model are approximately normally distributed with a given mean Mu = 43.8 miles per gallon and standard deviation Sigma = 5.1 miles per gallon. What percentage of the automobiles have an average miles per gallon between 38.7 miles per gallon and 48.9 miles per gallon?

A.) 68%

2. What is the mean of the normal distribution shown below?

C.) 4

3. The annual salaries of all employees at a financial company are normally distributed with a mean of $34,000 and a standard deviation of $4,000. What is the z-score of a company employee who makes an annual salary of $54,000?

D.) 5

4. The lengths of a lawn mower part are approximately normally distributed with a given mean Mu = 4 in. and standard deviation Sigma = 0.2 in. What percentage of the parts will have lengths between 3.8 in. and 4.2 in.?

B.) 68%

5. The mean of the temperatures in the chart is 24° with a standard deviation of 4°. Which temperature is within one standard deviation of the mean?

C.) 22°

6. Which statement must be true?

D.) Each distribution has the same mean and a different standard deviation.

7. The heights of a certain type of tree are approximately normally distributed with a mean height Mu = 5 ft and a standard deviation Sigma = 0.4 ft. Which statement must be true?

D.) A tree with a height of 6.2 ft is 3 standard deviations above the mean.

8. What is the difference of the means of the distributions?

A.) 15

9. )The weights of boxes of candies produced in a factory are normally distributed with a mean weight of 16 oz and a standard deviation of 1 oz. What is the weight of a box of candies with a z-score of 2?

B.) 18 oz

10. What is the mean of the normal distribution shown below?

B.) 0

Charra [1.4K]3 years ago
5 0

Answer:

68 %

Step-by-step explanation:

Let's call X to the random variable ''lengths of a lawn mower part''.

X ~ N(given mean;standard deviation)

X ~ N(4 in;0.2 in)

To find the percentage,first we are going to turn this random variable X into a random variable Z. Z will be a N(0;1)

We do this by subtracting the given mean to the original variable and dividing by it deviation.

P (3.8 in < X < 4.2 in) =

P [(3.8 in - 4 in) / (0.2 in)] < [(X - 4 in)] / (0.2 in) < [(4.2 in - 4 in) / (0.2 in)]

P ( -1 < Z < 1 ) = P (Z < 1) - P (Z< -1)

Where we can find  P (Z < 1) and  P (Z< -1) in any table of a N(0;1)

P (Z < 1) - P (Z< -1) = 0.8413 - 0.1587 = 0.6826 = 68%

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Answer:

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

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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

\hat p=\frac{125}{180}=0.694 estimated proportion of americans between 17 to 24 that not qualify for the military

p_o=0.75 is the value that we want to test  

\alpha=0.05 represent the significance level  

Confidence=95% or 0.95  

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 less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :  

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Alternative hypothesis:p < 0.75  

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

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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{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

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 \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

If we compare the p value obtained and 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 the proportion of americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

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