How to calculate inverse cumulative distribution using a table?

Question

I need help with this: $$P(X\geq a)=1-F_X(a)=1-\Phi\left(\frac{a-70}{8}\right)=0.25$$

When $X\sim N(70,64)$.

I know that it should be: $(a-70)/8 = 0.6745$

How do I get $0.6745$ From $Z$ table? I know that $0.75$ is between $0.67$ and $0.68$ in that table. Do I need to do an average $(0.67+0.68)/2$?

the final answer is $a = 75.396$

Answer 1

So I am assuming that the original question is $$P(X\geq a) = .75$$

So, I agree, a next good step is $$1-P(X\leq a) = .75$$

This gives $$P(X\leq a) = .25$$

Standardization gives $$P\left(Z\leq \frac{a-70}{\sqrt{64}}\right) = .25$$

Then $$\frac{a-70}{8} = \Phi^{-1}(.25)$$

To get $\Phi^{-1}(.25)$, it depends on the table that you have and what your instructor told you to do. I usually hear find the closest value.

Many tables are defined differently. Here is one I found online. I looked for the closest value to $.25$ that I could find. I matched up the row and column and it gives me a z-score of $-.67$ (using a calculator gives $-0.6744898$).

So I have that $$\frac{a-70}{8} = -.67$$

This gives $a = 64.64$, which is wrong according the the answer given. One possible correction is to consider that the original question is actually $$P(X\leq a) = .75$$

By symmetry, $\Phi^{-1}(.25) = -.67$ implies $\Phi^{-1}(.75) = .67$ and so

$$\frac{a-70}{8} = .67$$ which gives $a = 75.36$

Answer 2

My table, which may be very similar to yours, gives $\Pr(Z\le 0.67)\approx 0.7486$ and $\Pr(Z\le 0.68)\approx 0.7517$.

To get a better approximation to the $z$ such that $\Pr(Z\le z)=0.75$, we do a linear interpolation.

The number $0.7486$ is $0.0014$ short of $0.75$, while $0.7517$ is $0.0017$ above $0.75$. So a better estimate for the appropriate $z$ is $$0.67+\frac{0.0014}{0.0014+0.0017}(0.68-0.67).\tag{1}$$ A calculator shows that $\frac{0.0014}{0.0031}\approx 0.45$. Then using (1) we get the correction you quoted.

This linear interpolation is a kind of weighted averaging. Since $0.0014$ and $0.0017$ are fairly close to each other, the weighted averaging gives a result quite close to the simple averaging you suggested.

Remark: In the bad old days, when table use was common for many things, such interpolations were a standard feature in calculations.