How to compute approximate probability with z value?

Question

This is the question:

An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable x with a mean value of 48 lb and a standard deviation of 18 lb. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With n = 100, the total weight exceeds the limit when the average weight x exceeds 6000/100.) (Round your answer to four decimal places.)

From my calculation I get a z value of 6.66 which would be 100%. Its being marked wrong by my teacher and I cannot figure out why. Every similar example I get a realistic z score and can find the value. But 48-60 is -12 18/(sqrt(100)= 1.8 -12/1.8 is -6.66.

Answer 1

We want to calculate the probability that the population mean exceeds $\frac{6000}{100}=60$

$P(x>60)=P(\frac{x-\mu}{\sigma/\sqrt{n}}>\frac{60-\mu}{\sigma/\sqrt{n}})=P(\frac{x-\mu}{\sigma/\sqrt{n}}>\frac{60-48}{18/\sqrt{100}})=P(z>\frac{60-48}{18/\sqrt{100}})=P(z>6.667)=1-\Phi (6.667)=\Phi (-6.667)$,

Note: z ~ N(0,1).

So, by using the R command $1-pnorm(\frac{60-48}{18/\sqrt{100}})$, we get 0.

So it is very unlikely that the weight of a single baggage will exceed 60, which means that the total weight is unlikely to exceed 6000.

Answer 2

Let $X_1, X_2, \ldots, X_{100}$ represent the random baggage weights of each of the $n = 100$ passengers. Then for each $i = 1, \ldots, 100$, each $X_i$ might be assumed to follow a normal distribution with mean $\mu = 48$ and standard deviation $\sigma = 18$. If these weights are all independent, then their sum $$S = X_1 + X_2 + \cdots + X_{100} \sim \operatorname{Normal}\left(\mu_S = 100\mu = 4800, \sigma_S = \sigma\sqrt{100} = 180\right).$$ That is to say, the the mean of the total weight is $4800$ with a standard deviation of $180$. Note here that I'm working with the total weight, not the average weight. So the probability that the total weight exceeds the limit of $6000$ is $$\Pr[S > 6000] = \Pr\left[\frac{S - \mu_S}{\sigma_S} > \frac{6000 - 4800}{180}\right] = \Pr[Z > 6.667].$$ Note the direction of the inequality: This probability corresponds to finding the probability that a standard normal random variable $Z$ is more than 6.667 standard deviations greater than zero. Thus, the probability will be very, very tiny--close to zero, rather than close to $1$. We would write $$\Pr[Z > 6.667] = 1 - \Phi(6.667) \approx 1.3083924686 \times 10^{-11},$$ in effect, it is less than a $1$ in $76$ billion chance.