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There are 80 terms, measured in seconds. The standard deviation is 77.96433 (sec). The mean is 267.2375 (sec).

What percentage are over 180 seconds? What percentage are between 210 seconds and 300 seconds? Under what length are 90% of the terms?

(I have no idea how to calculate percentages using the standard deviation and mean, that's where I'm stuck. The textbook I got doesn't explain it properly.)

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    Congratulations that you made it to here! That was a hard task. To help you on your journy, we need to know where you are stuck! So add this little detail, please. Thank you very much. – 2017-01-24
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    Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 2017-01-24
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    Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 2017-01-24
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    Do you know what the [standard deviation](https://en.wikipedia.org/wiki/Standard_deviation) and the [mean](https://en.wikipedia.org/wiki/Mean) are? – 2017-01-24

1 Answers 1

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The wording of the problem seems a little strange to me; not surprised you're confused.

I'm betting you have just started to learn about the normal distribution and that you are expected to assume that the 'terms' are normally distributed. And then I guess you are supposed to use printed tables of the standard normal distribution to get the 'percentages' (probabilities).

I will leave it to you do work out the exact problem, but I will give you some clues how to get started. Then maybe you can edit at least some first steps toward the exact problem into your Question, and ask for further help as needed.

Suppose the mean is 270 and the standard deviation is 85. That is, the data come from the distribution $X \sim Norm(\mu = 270,\, \sigma = 85).$ But you won't find a table for that distribution in your book. You have to change this to a problem for the standard normal distribution $Z \sim Norm(\mu - 0,\, \sigma = 1).$ The conversion for $X$ to $Z$ is done with the 'standardization' formula $$Z = \frac{X - \mu}{\sigma},$$ which you should try to find in your book. (The material nearby the equation may help you with this and similar problems.)

Now for the question, what percentage is over 180 sec.

$$P(X > 180) = P\left(\frac{X - \mu}{\sigma} > \frac{180 - 270}{85} \right) = P(Z > -1.06).$$

I don't know how the standard normal table in your book is organized. This distribution is symmetrical about $0,$ so $P(Z > -1.06) = P(Z < 1.06).$ Follow the directions and examples in your book to get $P(Z > -1.06) = 0.8554.$ So almost 86% of the 'terms' exceed 180.

Below is a sketch of the standard normal density function. The total probability beneath the density curve is $1.$ The probability $0.8554$ is the area to the right of the red line and beneath the curve. [The area between the vertical red line and the vertical green line is 0.3554. Exactly half (0.5000) of the area under the curve lies on either side of the vertical green line.]

enter image description here

Now try using the actual numbers in your first question as posted. Let us know what result you get when you transform $X = 180$ to the z-scale (instead of my -1.06), and what final answer you get (something like 87%), and describe any difficulties you have using the printed standard normal table. If you show some details, maybe we can give more help as required.