Is Z in standardized normal distribution always positive?

Question

The question asks me to find the mean, given that:

• $\sigma$ is $0.8$ when not standardized

• 96% is over 40

I worked out that $Z = -1.75$ in this case, which then would lead me to

$$Z = (\mu-40)/\sigma$$

but in the answers, it shows $1.75=(\mu-40)/\sigma$ instead of $-1.75$.

Would someone please be able to tell me why?

Should I always use the positive when doing similar questions?

Answer 1

The standard normal distribution is symmetric. And therefore

$P(X\geq 40)=1-\Phi \left( \frac{40-\mu}{0.8} \right)=\Phi \left( \frac{\mu-40}{0.8} \right)=0.96$

$\frac{\mu-40}{0.8}=\Phi^{-1}(0.96)$

$\frac{\mu-40}{0.8}=1.75$

To get your solution you multiply both sides by $(-1)$:

$\frac{40-\mu}{0.8}=-1.75$

Answer 2

Your question posits the existence of a normally distributed random variable $X$ with standard deviation $\sigma = 0.8$, for which $\Pr[X > 40] = 0.96$; that is, to say, the chance that $X$ is above $40$ is $96\%$. That is what those two conditions mean.

From this, we are supposed to determine the mean $\mu$ of $X$. To do this, we can standardize $X$: that is to say, the random variable $$Z = \frac{X - \mu}{\sigma} \sim \operatorname{Normal}(0, 1).$$ So, from the probability condition, we would write $$0.96 = \Pr[X > 40] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{40 - \mu}{0.8}\right] = \Pr\left[Z > \frac{40 - \mu}{0.8}\right].$$ Now, let's say that $$z^* = \frac{40-\mu}{0.8}.$$ This value of $z^*$ is called a z-score, and the condition $$0.96 = \Pr[Z > z^*]$$ means that $z^*$ is some real number for which 96% of the area under the standard normal density will be to the right of this value: see the figure.

So, if we study the animation, we can see that $z^*$ should be negative, and upon more careful inspection, it looks like it should be around $-1.7$. If we use a normal distribution table or a calculator, we would find that $z^* \approx -1.75069$. (On a TI-83 calculator, this is accomplished by entering invNorm(1 - 0.96), the reason being that invNorm(p) finds $z^*$ satisfying $\Pr[Z \le z^*] = p$, whereas we want to take the complementary probability).

So, now that we know that $z^* \approx -1.75069$, it is a simple matter to compute $$\mu = 40 - z^* \sigma \approx 38.5995.$$ This is the mean of the original distribution $X$.