# Distribution Probability

A magazine has a set limit of 3000 on the word count of its articles. The editor has noticed that the articles sent for publication are consistently over the limit and she estimated that the distribution of the word counts is $$N(3500, \sigma^2)$$.

a) Find $$\sigma^2$$ if only 20% of the submitted articles meet the word limit of 3000.

b) Determine what the word limit should be if we want 90% of the submitted articles to meet it.

c) What is the proportion of submitted articles with word counts between 3100 and 3300?

Any help or hints would be appreciated! Thanks!

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Since this may be homework, could you show us what you have tried and where you are having problems. – Henry Dec 14 '12 at 21:20

## 1 Answer

HINT: The probability of a randomly chosen submission meeting the word count of 3000 or less is 20%.

This probability can be computed using the cumulative distribution function:

$$P(0 \le X \le 3000) = .20$$ $$P(0 \le X \le 3000) = \int_{0}^{3000} f(x)\ dx = \int_{-\infty}^{3000} f(x)\ dx - \int_{-\infty}^0 f(x)\ dx$$

where $f(x)$ is the probability distribution function of the normally distributed random variable $X \sim \mathcal{N}(3500,\sigma^2)$.

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Your next step will to be to transform the distribution $\mathcal{N}(3500,\sigma^2)$ to the standard normal distribution, use this in the integration by change of variables, keeping in mind that your limits of integration must change. – Emily Dec 14 '12 at 21:50