β οΈ The calculator's invNorm works on the standard normal $Z$, so we must convert $\hat{P}$ into $Z$ first.
π€ Where Students Often Get Stuck
"Why do we subtract $\mu$ and divide by $\sigma$? Where does this formula come from?"
Many students treat $Z = \dfrac{X-\mu}{\sigma}$ as a formula to memorise.
In fact, it is just two geometric moves:
Step 1 β Subtract $\mu$ β shift the centre to 0
Step 2 β Divide by $\sigma$ β change units to "number of standard deviations"
π‘ By the end of these slides, $Z$ will not look like a formula. It will be the same data described in another way.
π― Intuition 1 β Subtract $\mu$ to Shift
The original distribution $X \sim N(\mu, \sigma^2)$ has its bell centred at $\mu$.
After shifting, the shape is unchanged; only the centre moved from $\mu$ to $0$.
The standard deviation is still $\sigma$.
π Intuition 2 β Divide by $\sigma$ to Rescale
The bell is still "wide or narrow" depending on $\sigma$. We want every normal distribution to map to one universal curve.
After dividing by $\sigma$, the horizontal unit changes from "$X$ units" to "numbers of standard deviations".
A value of $z = 1$ means "exactly one standard deviation above the mean".
π The Key Insight β A New "Ruler"
Imagine two students sitting different exams:
Student
Score $X$
Class $\mu$
Class $\sigma$
$Z = (X-\mu)/\sigma$
Alice
85
70
10
+1.5
Bob
62
50
8
+1.5
Their raw scores are very different (85 vs 62), but their $Z$ scores match.
Both students are "1.5 standard deviations above their class mean" β the same relative position.
π Standardising lets us compare different normal distributions on a common scale.
π The Formula and Probability Invariance
$$Z = \frac{X - \mu}{\sigma} \;\sim\; N(0,\,1)$$
Translating probability statements
Because standardising only changes the ruler, not the event:
$$\Pr(X < a) \;=\; \Pr\!\left(Z < \frac{a-\mu}{\sigma}\right)$$
The left side asks "the probability that $X$ falls below $a$".
The right side asks "the probability that $Z$ falls below $\tfrac{a-\mu}{\sigma}$".
They describe the same event in two coordinate systems.
This is precisely why a single standard normal table β or a single calculator function β solves any normal problem.
π Back to Example 12 β Standardising the Statement
We want $n$ such that $\Pr(\hat{P} < 0.58) = 0.3446$.
Since $0.3446 < 0.5$, the corresponding $z$ value is negative (left of the mean).
How to call invNorm
menu β 5: Probability
β 5: Distributions
β 3: Inverse Normal
Area: 0.3446 ΞΌ: 0 Ο: 1
β z β β0.400
What does $-0.400$ mean?
It is the standard-normal quantile that satisfies $\Pr(Z < z) = 0.3446$.
So the two sides agree:
$$\frac{-0.02}{\sqrt{0.24/n}} = -0.400$$
π‘ Rule of thumb: any inverse-probability lookup is always done on the standard normal $N(0,1)$.
βοΈ Solving the Equation for $n$
1Start from $\dfrac{-0.02}{\sqrt{0.24/n}} = -0.400$ and multiply both sides by the square root:
$$-0.02 = -0.400 \cdot \sqrt{\frac{0.24}{n}}$$
2Divide both sides by $-0.400$:
$$\sqrt{\frac{0.24}{n}} = 0.05$$
3Square both sides:
$$\frac{0.24}{n} = 0.0025$$
4Solve for $n$:
$$n = \frac{0.24}{0.0025} = \boxed{96}$$
β The required sample size is 96 people.
πΊοΈ The Whole Roadmap
1 Model $\hat{P} \sim N(\mu,\sigma^2)$ $\mu=p,\ \sigma=\sqrt{\tfrac{p(1-p)}{n}}$
2 Standardise Rewrite as $\Pr(Z < z) = \text{given}$ by translating $\hat{P}$ to $Z$
3 Inverse $z$ Use invNorm(area, 0, 1) to recover the matching $z$
4 Substitute back $z = \dfrac{a-\mu}{\sigma}$ becomes an equation in $n$
5 Solve algebraically Rearrange and find $n$
6 Sanity check Is $n$ large enough for the normal approximation?
π― Standardising is the bridge from the question's $X$ or $\hat{P}$ to the calculator's $Z$.
Once "shift by $\mu$, scale by $\sigma$" feels natural, the formula stops being something to memorise.
π Practice Problem
Question. Suppose 80% of students hand in homework on time. From a random sample of $n$ students,
$\Pr(\hat{P} > 0.85) = 0.1056$. Find $n$.