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3 years ago

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When looking for a job micah is told that the monthly salary has a range of 876 he also knows that the starting salary is 1,793

per month what is the maximum amount of money that micah can earn in this position?

1 answer:

12345 [234]3 years ago

6 0

Answer:

2669

Step-by-step explanation:

Given that :

Salary range = 876

Starting salary = 1793 monthly

The starting salary represents the minimum amount Micah can earn monthly

Therefore, the amount of money much can earn can be calculated thus :

Recall :

Range = maximum - minimum

876 = maximum - 1793

876 + 1793 = maximum

Maximum = 2669

Hence, the maximum salary Micha can earn is 2669 monthly

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Step-by-step explanation:

The area of a triangle = 1/2bh

where b is the base and h is the height

by substitution:

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SashulF [63]

Answer:

a)For this case the best estimator for the true mean is the sample mean $\hat \mu =\bar X$ because:

$E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}$

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$E(\bar X)=\frac{1}{n} n\mu =\mu$

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Step-by-step explanation:

Previous concepts

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

$\bar X=4.1$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=0.4$ represent the population standard deviation

n=50 represent the sample size

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=0.4$

The sample mean $\bar X$ is distributed on this way:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

A. What is the point estimate in this scenario? Is it an unbiased estimator?

For this case the best estimator for the true mean is the sample mean $\hat \mu =\bar X$ because:

$E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}$

And if we assume that each observation $X_1 , X_2,...,X_n$ follows a normal distribution $X_i \sim N(\mu,\sigma)$ then we have:

$E(\bar X)=\frac{1}{n} n\mu =\mu$

So then yes the $\bar X[/tex[ is a unbiased estimator for the true mean. B. What is the critical z-value for a 95% interval?
The next step would be find the value of [tex]\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=1.96$

C. What is the margin of error for a 95% interval in this scenario? Show your work.

The margin of error is given by:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

If we replace the values that we have we got:

$ME= 1.96 \frac{0.4}{\sqrt{50}}=0.1109$

The confidence interval on this case is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since we have all the values we can replace:

$4.1 - 1.96\frac{0.4}{\sqrt{50}}=3.989$

$4.1 + 1.96\frac{0.4}{\sqrt{50}}=4.211$

So on this case the 95% confidence interval would be given by (3.989;4.211)

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Answer:

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Step-by-step explanation:

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Now, all we have to do is plug in out x values in order to get our y values and complete the table.

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professor190 [17]

In this shape there will be 540 degrees

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Answer:

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Step-by-step explanation:

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