Question

The restaurant in a large commercial building provides coffee for the occupants in the building. The restaurateur has determined that the mean number of cups of coffee consumed in a day by all the occupants is \(2.0\) with a standard deviation of .6. A new tenant of the building intends to have a total of 125 new employees. What is the probability that the new employees will consume more than 240 cups per day?

Answer 1

The objective is to find the probability that the new employees will consume more than 240 cups per day.

Let \(X\) denote the number of cups of coffees consumed per person and it follows the normal distribution with mean of \(\mu=2.0\) cups and standard deviation of \(\sigma=0.6\) cups

The total number of new employees, \(n=125\)

Let the random variable \(T\) denote the total number of cups of coffees consumed by the new employees.

The mean and standard deviation of total number of cups of coffee consumed by 125 employees is,

\(\begin{aligned} E(T) &=\mu_{T} \\ &=n \mu \\ &=125 \times \mu \\ &=125 \times 2.0 \\ &=250 \\ \sigma_{T}=& \sqrt{n} \times \sigma \\=& \sqrt{125} \times 0.6 \\=& 6.71 \end{aligned}\)

Hence, the random variable \(T\) follows the normal distribution with mean of 250 and \(6.71\). That

is, \(T \quad N\left(\mu_{T}=250, \sigma_{T}=6.71\right)\).

Note: We are supposed to find the probability that the new employees will consume more than 240 cups per day. In other words, we are supposed to find the probability the total number of cups consumed by 125 employees is more than 240 cups.

Now, the total number of cups consumed by 125 employees is more than 240 cups is,

$$ \begin{aligned} P(T>240) &=1-P(T \leq 240) \\ &=1-P\left(\frac{T-\mu_{T}}{\sigma_{T}} \leq \frac{240-250}{6.71}\right) \\ &=1-P(Z \leq-1.49) \\ &=1-0.0681 \\ &=0.9319 \end{aligned} $$

Therefore, there is a \(93.19 \%\) of chance that the new employees will consume more than 240 cups per day.