A set of data has a normal distribution with a mean of 180 and a standard deviation of 20. What percent of the data is in the interval 140 - 220?

allucinemsj
2022-08-13
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Macie Melton

Answered 2022-08-14
Author has **19** answers

We are given the information that this distribution is normal with a mean $\mu $ of 180 and a standard deviation $\sigma $ of 20. This describes a distribution $N(180,{20}^{2})$.

To answer the question, we will convert this problem into a standard normal distribution ${N}_{s}(0,{1}^{2})$ question by determining the z-scores for the interval 140-220. This can be done either by "eyeballing it" (since $\sigma $ is 20, and each endpoint of the interval is a multiple of $\sigma $ away from the mean $\mu $), or we can use the z-score formula:

$z=\frac{x-\mu}{\sigma}$

Thus:

${z}_{140}=\frac{140-180}{20}=-\frac{40}{20}=-2\phantom{\rule{0ex}{0ex}}{z}_{140}=\frac{220-180}{20}=\frac{40}{20}=2\phantom{\rule{0ex}{0ex}}$

This tells us the interval we're being asked about is analogous to determining what percent of the standard normal distribution ${N}_{s}$ lies between z-scores of -2 and 2.

In statistics there is a handy "rule of thumb" sometimes called the Empirical Rule which says the approximately 95% of the data in a normal distribution lies in the interval $[-2\sigma ,2\sigma ]$, which is exactly what we're being asked. (The actual answer is more like 95.45%.)

To answer the question, we will convert this problem into a standard normal distribution ${N}_{s}(0,{1}^{2})$ question by determining the z-scores for the interval 140-220. This can be done either by "eyeballing it" (since $\sigma $ is 20, and each endpoint of the interval is a multiple of $\sigma $ away from the mean $\mu $), or we can use the z-score formula:

$z=\frac{x-\mu}{\sigma}$

Thus:

${z}_{140}=\frac{140-180}{20}=-\frac{40}{20}=-2\phantom{\rule{0ex}{0ex}}{z}_{140}=\frac{220-180}{20}=\frac{40}{20}=2\phantom{\rule{0ex}{0ex}}$

This tells us the interval we're being asked about is analogous to determining what percent of the standard normal distribution ${N}_{s}$ lies between z-scores of -2 and 2.

In statistics there is a handy "rule of thumb" sometimes called the Empirical Rule which says the approximately 95% of the data in a normal distribution lies in the interval $[-2\sigma ,2\sigma ]$, which is exactly what we're being asked. (The actual answer is more like 95.45%.)

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So I have $X\sim Bin(350,\frac{13}{100})$

Then let $Y\sim N(\frac{13}{100},{\sqrt{0.00032314}}^{2})$

I want $P(X<40)=P(\hat{p}<\frac{4}{10})$

Then $P(z<\frac{\frac{4}{10}-\frac{13}{100}}{{\sqrt{0.00032314}}^{2}})$ which is obviously very wrong.

Where did it go wrong, and why did it go wrong?

So I have $X\sim Bin(350,\frac{13}{100})$

Then let $Y\sim N(\frac{13}{100},{\sqrt{0.00032314}}^{2})$

I want $P(X<40)=P(\hat{p}<\frac{4}{10})$

Then $P(z<\frac{\frac{4}{10}-\frac{13}{100}}{{\sqrt{0.00032314}}^{2}})$ which is obviously very wrong.

Where did it go wrong, and why did it go wrong?

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On average, $30\mathrm{\%}$ were further than ___ kilometers away when they had their accident.

Is $30\mathrm{\%}$ a z-score or is it a mean? Transport Canada was investigating accident records to find out how far from their residence people were to when they got into a traffic accident. They took the population of accident records from Ontario and measured the distance the drivers were from home when they had their accident in kilometers (km). The distribution of distances was normally shaped, with $\mu =30$ kilometers and $\sigma =8.0$ kilometers.

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