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Harlamova29_29 [7]

1 year ago

7

5. Kal Tire installs automobile tires on a first-come first-served basis. A random sample of 50 customers experienced an average

wait time of 93.7 minutes. Assume that the standard deviation of total wait time for all customers is 20.6 minutes. Determine the margin of error for a 95% confidence interval for this sample.

1 answer:

Rainbow [258]1 year ago

7 0

5.7099 is the margin of error.

<h3>What is standard deviation ?</h3>

A standard deviation (or σ) is a measure of how dispersed the data is in relation to the mean.

Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.

According to the question,

The standard sampling error of the sample mean is σₓ ,

The sampling distribution is: N (u,σₓ/n)

Therefore,

By using the standard deviation formula:

σₓ = √∑(xi -μ)/N

Where,

∑= population standard deviation

N= the size of the population

xi= each value from the population

μ =the population mean

So ,

σₓ = $\frac{20.6}{\sqrt{50} }$ = 2.913

Since, a = 1- 95% = 0.05

therefore $Z\frac{a}{2}$ = (1-0.005 x 2) = 1.959964

Here, the margin of error for a 95% confidence interval for this sample is given by:

$Z\frac{a}{2}$ * σₓ = 1.959964 x 2.913

= 5.7099

5.7099 is the margin of error for a 95% confidence interval for this sample.

Learn more about standard deviation here:

brainly.com/question/13905583

#SPJ1

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2 years ago

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PolarNik [594]

Use the formula 42600 ÷ (6.9 ÷<span> 100) to calculate this !

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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%20x%20%5E%7B3%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ - x ^{3} } " alt=

noname [10]

I'm guessing you're given the function $y(x)=2-x^3$ , and you're asked to find the inverse function $y^{-1}(x)$ . To do this, swap $x$ and $y$ , then solve for $y$ :

$x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}$

so that the inverse function is

$y^{-1}(x)=\sqrt[3]{2-x}$

Just to verify:

$y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x$

$y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x$

But in case you're actually only interested in computing the square root, first we note that $\sqrt x$ (the real-valued square root) is only defined as long as $x\ge0$ . So $\sqrt{-x^3}$ is defined as long as $-x^3\ge0$ , or $x^3\le0$ , or equivalently $x\le0$ . Under this condition, we could write

$\sqrt{-x^3}=\sqrt{-x\times x^2}=\sqrt{-x}\sqrt{x^2}$

We can simplify this further, but we have to be careful. Suppose $x=-1$ . Then $x^2=(-1)^2=1$ . But we get the same result if $x=1$ , since $x^2=1^2=1$ . There are two possible values of $x$ that given the same value of $x^2$ , so to capture both of them, we take $\sqrt{x^2}=|x|$ , the absolute value of $x$ . Then

$\sqrt{-x^3}=|x|\sqrt{-x}$

We can't simplify the square root term further than this.

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3 years ago