Question

Customers using a self-service soda dispenser take an average of 12 ounces of soda with an SD of 4 ounces. Assume that the amount would be normally distributed.

a. What is the chance that a randomly selected customer takes over 10 ounces of soda?
b. What is the chance that a randomly selected customer takes between 13 and 14 ounces of soda?
c. What is the chance that the next 100 customers will take an average of less than 12.24 ounces?

Answer 1

Part a). We asked to calculate $P(X > 10)$ . Let's change the variable

$$Z = \frac{{X - \mu }}{\sigma } = \frac{{X - 12}}{4}$$

Then $Z$ will be distributed according to standard normal distribution $Z \sim N(0,1)$. Let's rewrite $P(X > 10)$ as

$$P(X > 10) = P(Z > \frac{{10 - 12}}{4}) = P(Z > - \frac{1}{2}) = P(Z < \frac{1}{2})$$

(at the last one we use simmetry of the standard normal distribution about $x = 0$). Now let's use the following table

and get $P(Z < \frac{1}{2}) \approx 0.6915$, thus

$$P(X > 10) \approx 0.6915$$

Part b). We asked to calculate $P(13 < X < 14)$. Using the same variable $Z$ we can rewrite it as

$$P(13 < X < 14) = P(\frac{1}{4} < Z < \frac{1}{2}) = P(Z < \frac{1}{2}) - P(Z < \frac{1}{4})$$

Using the table $P(Z < \frac{1}{2}) \approx 0.6915$ and $P(Z < \frac{1}{4}) \approx 0.5987$ . Thus

$$P(13 < X < 14) \approx 0.6915 - 0.5987 \approx 0.0928$$

Part c). Let ${X_i}$ be the amount of ounces of soda that the i-th customer will get, and ${X_i} \sim N(12,4)$ . Let's use the following lemma: Let $\xi \sim N({\mu _1},{\sigma _1})$ and $\eta \sim N({\mu _2},{\sigma _2})$ be the two normally distributed random values. Then $(\xi + \eta ) \sim N({\mu _1} + {\mu _2},\sqrt {\sigma _1^2 + \sigma _2^2} )$ . You may find two different proofs of this here

https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

Using this lemma we see that

$$Y = \sum\limits_{k = 1}^{100} {{X_k}} \sim N(1200,40)$$

Now we need to find $P(\frac{Y}{{100}} < 12.24) = P(Y < 1224)$ . Let's change the variable

$$Z = \frac{{Y - 1200}}{{40}}$$

and we get

$$P(\frac{Y}{{100}} < 12.24) = P(Z < \frac{3}{5})$$

Using the table $P(Z < \frac{3}{5}) \approx 0.7257$ thus

$$P(\frac{Y}{{100}} < 12.24) \approx 0.7257$$

This is the answer.