Finding constant for CLT normal distribution

Question

(This is from my textbook, but I don't understand their explanation. I've Googled around, but haven't found an answer that makes sense.)

$$ \mu = 0, \sigma^2 = 1, n =16 $$

Find c such that:

$$ P(\bar{x} \leq c) = 0.75 $$

$$ \Rightarrow P \left(\frac{\bar{x}-0}{\sqrt{\frac{1}{16}}} \leq \frac{c-0}{\sqrt{\frac{1}{16}}} \right) $$

$$ \Rightarrow P(Z \leq 4c) $$

$$ \Rightarrow 4c = 0.675 $$

$$ \Rightarrow c = 0.169 $$

I understand switching to standard normal, but how do they pull out $4c = 0.675$?

I'm sure it's trivial, but I'm just not getting it.

Answer 1

You forgot to bring down the RHS of the equation. 0.675 according to the standard normal distribution table (z-table) is the number d s.t. P(z < d)=0.75

Then d = 4c.

Specifically, d is the upper bound satisfying:

$\int_{-\infty}^{d} f_Z(z) dz = 0.75$

where $Z$ is standard normal. This is computed using numerical methods I guess. Excel has a function called NORMSINV.

d would be equal to "NORMSINV(0.75)"

If so inclined, compute for yourself:

$\int_{-\infty}^{0.675} f_Z(z) dz$

Answer 2

Hint:

Normsinv(0.75) = 0.675. where normsinv is the inverse standard normal you can plug in the formula in EXCEL and get it.

Answer 3

You can find it by looking at a percentage points table for the normal distribution. The table gives you a $z$ score such that $P(Z\le z)=p$ for different values of $p$.

Your $p=0.75$

The $4c$ is just simplifying the $\frac{c}{\sqrt{\frac{1}{16}}}$