Answer to Question #98393 in Statistics and Probability for Tasnia Islam

Question

Customers using a self-service soda dispenser take an average of 12 ounces of soda with an SD of 2 ounces. Assume that the amount would be normally distributed.

What is the probability that a randomly selected customer takes over 10 ounces of soda?
What is the probability that a randomly selected customer takes between 13 to 14 ounces of soda?
How many of the next 100 customers will take an average of less than 12.24 ounces?

Answer 1

Let $X=$ the number of ounces of soda: $X\sim N(\mu, \sigma^2)$

Then

$$Z={X-\mu \over \sigma}\sim N(0,1)$$

Given that $\mu=12 \ ounces , \sigma=2\ ounces$

$$P(X>10)=1-P(X\leq 10)=1-P(Z\leq {10-12 \over 2})=$$

$$=1-P(Z\leq -1)\approx1-0.158655\approx0.8413$$

The probability that a randomly selected customer takes over 10 ounces of soda is $0.8413$

$$P(13<X<14)=P(X<14)-P(X<13)=$$

$$=P(Z< {14-12 \over 2})-P(Z< {13-12 \over 2})=$$

$$=P(Z< 1)-P(Z<0.5)\approx0.841348-0.691462\approx$$

$$\approx0.1499$$

The probability that a randomly selected customer takes between 13 to 14 ounces of soda is $0.1499$

Given that $n=100$

$$P(X<12.24)=P(Z<{12.24-12 \over 2/\sqrt{100}})=P(Z<1.2)\approx$$

$$\approx0.8849$$

$$0.8849\cdot100=88.49\approx88$$

88 of the next 100 customers will take an average of less than 12.24 ounces.