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Ne4ueva [31]
2 years ago
4

High-power experimental engines are being developed by the Stevens Motor Company for use in its new sports coupe. The engineers

have calculated the maximum horsepower for the engine to be 810HP. Sixteen engines are randomly selected for horsepower testing. The sample has an average maximum HP of 890 with a standard deviation of 85HP. Assume the population is normally distributed. Step 1 of 2 : Calculate a confidence interval for the average maximum HP for the experimental engine. Use a significance level of α=0.01. Round your answers to two decimal places.
Mathematics
1 answer:
____ [38]2 years ago
6 0

Answer:

The confidence interval for the average maximum HP for the experimental engine is between 639.53HP and 1140.47HP

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 16 - 1 = 15

0.01 significance level.

So a 1 - 0.01 = 0.99 = 99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 15 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.9467

The margin of error is:

M = T*s = 2.9467*85 = 250.47

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 890 - 250.47 = 639.53HP

The upper end of the interval is the sample mean added to M. So it is 890 + 250.47 = 1140.47HP

The confidence interval for the average maximum HP for the experimental engine is between 639.53HP and 1140.47HP

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Answer:

0% probability that at least 1,500 will agree to respond

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 15000, p = 0.09

So

\mu = E(X) = np = 15000*0.09 = 1350

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05

What is the probability that at least 1,500 will agree to respond

This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{1499 - 1350}{35.05}

Z = 4.25

Z = 4.25 has a pvalue of 1.

1 - 1 = 0

0% probability that at least 1,500 will agree to respond

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Answer:

hope that this picture answers it

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we have

10x^{2} + 40x - 13 = 0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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Complete the square. Remember to balance the equation by adding the same constants to each side

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10(x^{2} + 4x+4) =53

Rewrite as perfect squares

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Finding the mean;

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Outliers: brainly.com/question/12612191

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