Find the standard deviation $\sigma $ of a Normal Distribution

Question

The question is: Consider a normal distribution curve where the middle $70$ % of the area under the curve lies above the interval $( 8 , 20 )$. Use this information to find the mean and the standard deviation of the distribution.

I already found the mean using the range but the standard deviation I don't know what to do, could someone help me?

Thanks in advance.

Answer 1

Let $\mu, \sigma$ be the mean and standard deviation of the given distribution. The normal distribution is symmetric about the mean. Thus $\mu = \dfrac{8+20}{2} = 14$. Let $c$ be the point at $x = 20$, then the area from $\mu$ to $c$ is $0.35$, and $P(x < c) = 0.85\implies P\left(z < \dfrac{c - \mu}{\sigma}\right) = 0.85\implies \dfrac{c-\mu}{\sigma} \approx 1.04\implies \sigma \approx \dfrac{c - \mu}{1.04}=\dfrac{20-14}{1.04} = \dfrac{6}{1.04}= 5.77$ .

Answer 2

Let $X$ be such a random variable. If $\mu$ is the mean, and $\sigma$ the standard deviation then $$Y=\frac{X-\mu}{\sigma}$$ has mean $0$ and standard deviation $1$. Using for example wolfram alpha we find that for the standard normal distribution we have $70\%$ of the area on the interval $[-1.04,1.04]$. This means that $P(Y\in[-1.04,1.04])=0.7$. Furthermore $P(Y\in(-\infty,-1.04))=0.15$ and $P(Y\in(1.04,\infty))=0.15$. It follows that $$P(X\in[\mu-1.04\sigma,\mu+1.04\sigma])=0.7$$ $$P(X\in(-\infty,\mu-1.04\sigma))=0.15$$ $$P(X\in(\mu+1.04\sigma,\infty))=0.15$$ which means that the interval $[\mu-1.04\sigma,\mu+1.04\sigma]$ is exactly $70\%$ of the area, and is also "in the middle". We can now solve the equations $\mu-1.04\sigma=8$ and $\mu+1.04\sigma=20$ to get $\mu=14$ and $\sigma\approx5.769$.