Statistics: finding the mean of a normally distributed variable.

Question

For a normally distributed variable B variance is (0.714)^2 and mean is x.

If P( x < b < 3.2) = 0.475 find x.

My working:

P ( x < b < 3.2) = P ( b < 3.2 )- P ( b < * x* )

P ( b < 3.2 ) - P ( b < * x* ) = 0.475

P ( z < [ 3.2 - x / 0.714 ] ) - P ( z < 0 ) = 0.475

P ( z < [ 3.2 - x / 0.714 ] ) - 0.5000 = 0.475

P ( z < [ 3.2 - x / 0.714 ] ) = 0.975

phi ( [ 3.2 - x / 0.714 ] ) = phi (1.960)

3.2 - x / 0.714 = 1.960

x = 1.8005

However this is incorrect, it should be 4.60. In the second to last line if 1.960 is negative this is what you get.

What did I do wrong so that 1.960 is positiv instead of negative?

Answer 1

Your result is right. You can check it by using your result. $B$ is disributed as

$B\sim\mathcal N(1.8005,0.714^2)$. Thus

$$P ( 1.8005 < B < 3.2)=\Phi\left( \frac{3.2-1.8005}{0.714} \right)-\Phi\left( \frac{1.8005-1.8005}{0.714} \right)$$

$$\Phi\left( 1.96 \right)-\Phi\left( 0 \right)=0.975-0.5=0.475$$

---

You get the result of $\mu=4.6 \ $ if $$\color{blue}{P( 3.2 < B < x^*) = 0.475}$$

$$\Phi\left( \frac{x^*-x^*}{0.714} \right)-\Phi\left( \frac{3.2-x^*}{0.714} \right)=0.475$$

$$0.5-\Phi\left( \frac{3.2-x^*}{0.714} \right)=0.475$$

$$-\Phi\left( \frac{3.2-x^*}{0.714} \right)=-0.025$$

$$ \frac{3.2-x^*}{0.714} =\Phi^{-1}(0.025)$$

$$ \frac{3.2-x^*}{0.714} =-1.96$$

$$\Rightarrow x^*=\mu=4.59974\approx 4.6$$