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Drag the values to the correct locations on the image. Not all values will be used.
Susan is conducting a survey about the electric bills of households in her city. From the electric company, she found that the population mean is
$98.75 with a standard deviation of $10.45. Susan has a sample size of 60.
Complete the equation that Susan can use to find the interval in which she can be 68% sure that the sample mean will lie.
98.75
109.2
88.3
7.75
60
10.45
C
+
r

Drag the values to the correct locations on the image Not all values will be used Susan is conducting a survey about the electric bills of households in her cit class=

Respuesta :

Answer:

[tex]x\\[/tex] ± [tex]\frac{s}{\sqrt{n}}[/tex] = [tex]9.78[/tex] ± [tex]\frac{10.45}{\sqrt{60}}[/tex]

Step-by-step explanation:

x is the value of our population/sample mean. Given that it's 98.75, we substitute x for 98.75.

We follow this pattern for s and n, which are the standard deviation and sample size respectively, to fill out the rest of our equation. This means that s is 10.45 and n is 60

Since we know the formula is [tex]x\\[/tex] ± [tex]\frac{s}{\sqrt{n}}[/tex], we can input our new values.

This gives us our answer of [tex]9.78[/tex] ± [tex]\frac{10.45}{\sqrt{60}}[/tex]

TL;DR

I took the test and the answer was right

The interval in which the population mean $98.75 with standard deviation of $10.45 with a sample size of 60 is given by its lower and upper limit as given by: 98.75 ± 10.45/(√(60))

What is empirical rule?

According to the empirical rule, also known as 68-95-99.7 rule, the percentage of values that lie within an interval with 68%, 95% and 99.7% of the values lies within one, two or three standard deviations of the mean of the distribution.

[tex]P(\mu - \sigma < X < \mu + \sigma) \approx 68\%\\P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 95\%\\P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 99.7\%[/tex]

where we had  where mean of distribution of X is [tex]\mu[/tex]  and standard deviation from mean of distribution of X is [tex]\sigma[/tex] (assuming X is normally distributed).

What is the distribution of sample mean for large samples?

Suppose X has has a distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], then, the distribution of the mean of samples of size n, drawn from the values of X, is normally distributed with mean equal to the mean of X, and standard deviation equal to [tex]\sigma/\sqrt{n}[/tex]

Symbolically, we write it as:

[tex]X \sim N(\mu, \sigma) \implies \overline{X} \sim N(\mu,\sigma/\sqrt{n} )[/tex]

(for large enough samples, n > 30)

For this case, we're specified that:

  • Population mean = [tex]\mu = $98.75[/tex] (dollars)
  • Population standard deviation = [tex]\sigma = 10.45[/tex] (dollars)
  • sample size = n = 60

Then, the distribution of the sample mean is approximately normal distribution with mean $98.75 and standard deviation approximately [tex]10.45/\sqrt{60}[/tex]

Symbolically, we get:

[tex]X \sim N(98.75, 10.45/\sqrt{60})[/tex]

Then, using empirical rule, we find the limits for sample mean for which we're 68% sure that the sample mean will lie in that interval as one standard deviation (of the sample mean) away from its mean.

[tex]P\left( 98.75 - 10.45/\sqrt{60} < \overline{X} < 98.75 + 10.45/\sqrt{60}\right) \approx 68\%[/tex]

So, limits are:

[tex]98.75 \pm \dfrac{10.45}{\sqrt{60}}[/tex]

Thus, the interval in which the population mean $98.75 with standard deviation of $10.45 with a sample size of 60 is given by its lower and upper limit as given by: 98.75 ± 10.45/(√(60))

Learn more about empirical rule here:

https://brainly.com/question/13676793

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