statistics: probability, normal distribution

Question

The time that customers take to complete their transaction at a money machine is a random variable with mean $\mu$ = $2$ minutes and standard deviation $\sigma$ = $0.6$ minutes.

About 30% of customers take more than 3 minutes to complete their transaction. Take a random sample of size $50$.

Find the probability that the selected sample takes on average between 1.8 minutes and 2.25 minutes.

I tried----

First when I've read the question I was thinking I need to use central limit theorem so I did

$$n = 50\\ \sigma = 0.6 \\ \mu = 2$$

$\mathrm{P}(1.8 < X < 2.25)$

then applied CLT:

$\displaystyle\mathrm{P}\left(\frac{1.8 - \mu}{\sigma/n^{1/2}} < X < \frac{2.25 - \mu}{ \sigma/n^{1/2}} \right) $

and I was going to just plug in the given values..

But then I'm confused about

"About 30% of customers take more than 3 minutes to complete their transaction.. "

How should I apply this with CLT ?? Is this mean 30 % of 50? so 15 customers are taking more than 3 minutes ?? So instead using 50, I should use 15 ??

Answer 1

The $z$-score of $0.3$ is $-0.52$. Consider that

$$0.52 \times 0.6 = 0.31$$ and

$$2 + 0.31 = 2.31$$

It is clear that the distribution is not normal. In fact, it is right-tailed.

I will say that since the sample size is large, consider using CLT because the distribution of the sample is normal, despite the distribution of the population.

To use CLT, $n=50$, $\mu = 2$, $\sigma = 0.6$. One thing I learnt constantly in my math class is to define the probability distribution. So in this case, I will say let $X$ be the distribution of the sample of time taken to complete their transaction. $$ P(1.8<X<2.5) = P(\frac{1.8-2}{\frac{\sqrt{0.6}}{50}}\leq X \leq \frac{2.5-2}{\frac{\sqrt{0.6}}{50}}) $$