Physics, 08.04.2020 23:29 mbalderp5bxjo

The mean number of words per minute (WPM) typed by a speed typist is 73 with a variance of 144. What is the probability that the sample mean would differ from the population mean by greater than 0.6 WPM if 77 speed typists are randomly selected? Round your answer to four decimal places.

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Answer from: Calirose

$P(73-0.6 < \bar X< 73+0.6)$

and using the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

We got:

$z = \frac{72.4-73}{\frac{12}{\sqrt{77}}}= -0.439$

$z = \frac{73.6-73}{\frac{12}{\sqrt{77}}}= 0.439$

And usint the normal standard distribution or excel we got:

P(-0.439< Z

And using the complement rule we got the probability desired "that the sample mean would differ from the population mean by greater than 0.6 WPM "

P= 1-0.3394=0.6606

Explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the number of words per minute typed of a population, and for this case we know the following info:

Where $\mu=73$ and $\sigma=\sqrt{144}=12$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

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

For this case the sample size is large enough n = 77>30 to apply the central limit theorem.

For this case we can begin calculating this probability:

$P(73-0.6 < \bar X< 73+0.6)$

And using the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

We got:

$z = \frac{72.4-73}{\frac{12}{\sqrt{77}}}= -0.439$

$z = \frac{73.6-73}{\frac{12}{\sqrt{77}}}= 0.439$

And usint the normal standard distribution or excel we got:

P(-0.439< Z

And using the complement rule we got the probability desired "that the sample mean would differ from the population mean by greater than 0.6 WPM "

P= 1-0.3394=0.6606

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The mean number of words per minute (WPM) typed by a speed typist is 73 with a variance of 144. What...

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