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3 years ago

What number is halfway between one sixth and five ninth

1 answer:

5 0

$\frac{\frac{1}{6}+\frac{5}{9}}{2}=\\
\frac{\frac{9}{54}+\frac{30}{54}}{2}=\\
\frac{\frac{39}{54}}{2}=\\
\frac{39}{108}$

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Nadusha1986 [10]

Answer:

$n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92$

So the answer for this case would be n=92 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

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".

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

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

And on this case we have that ME =1.5 and we are interested in order to find the value of n, if we solve n from equation (b) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 92% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.04;0;1)", and we got $z_{\alpha/2}=1.75$ , replacing into formula (b) we got:

$n=(\frac{1.75(8.2)}{1.5})^2 =91.52 \approx 92$

So the answer for this case would be n=92 rounded up to the nearest integer

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3 years ago