Using standardized normal distribution to find probability

Question

"The number of nails in a carton is $N(563.3, 33.2)$. The distributor guarantees that the number of nails is more than $n$ with a probability of 99%. Decide n."

I've tried to make this into the standardized normal distribution with $P(Z >\frac{n-563.3}{33.2}) = 0.99$ but I don't know where to take it from there, or if this even is a correct way to tackle this problem. Some hints and help would be greatly appreciated.

Answer 1

It is correct way. You need corresponding quantile for normal distribution - ie such value $x$ that $P(Z > x) = 0.99$. It's probably assumed that you take it from table (note that there are two-sided quantiles - $0.98$ two-sided quantile is equal to $0.99$ one-sided) or from some calculator like wolfram.

Answer 2

Let $X$ represent the random number of nails in a carton. Then, when the distributor guarantees that the probability there are more than $n$ nails in a randomly selected box is $0.99$, this means $$\Pr[X > n] = 0.99.$$ Standardizing, we get $$\Pr\left[\frac{X - \mu}{\sigma} > \frac{n - 563.3}{33.2}\right] = 0.99.$$ What is the value of $z^*$ corresponding to $\Pr[Z > z^*] = 0.99$? Right away, you should note that $z^* < 0$, which means that $n$ should be less than the mean value $\mu = 563.3$. This makes sense because if $n = \mu$, then there would be only a $0.5$ probability of a randomly selected box having more than $n$ nails. The smaller $n$ is, the more likely you'll get a box with more than $n$ nails.

Answer 3

Let $X\sim \mathcal N(\underbrace{563.3}_{\mu}, \underbrace{33.2}_{\sigma^2})$.

The probability that the number of nails is more than $n$ is: $$\mathbb P(X>n)=0.99 \iff \mathbb P(X-\mu>n-\mu)=0.99 \iff \\ \mathbb P\left(\frac{X-\mu}{\sigma}>\frac{n-\mu}{\sigma}\right)=0.99 \iff \\ P\left(Z>\frac{n-563.3}{\sqrt{33.2}}\right)=0.99 \iff P(Z>-2.3263479)=0.99 \Rightarrow \\ \frac{n-563.3}{\sqrt{33.2}}=-2.3263479 \Rightarrow n\approx 549.9$$ Now, let's see the neighborhood of $n=549.9$: $$\mathbb P\left(Z>\frac{549-563.3}{\sqrt{33.2}}\right)=0.9935;\\ \mathbb P\left(Z>\frac{550-563.3}{\sqrt{33.2}}\right)=0.9895.$$ Hence, the smaller $n$, the higher the probability. So, for $n\le 549$, the probability of having more than $n$ nails is at least $0.99$.