Question

The manager of a local soft drink bottling company believes that a new beverage- dispensing machine, which is designed to dispense 7 ounces, in fact dispenses anywhere between 6.5 and 7.5 ounces.
If the amount dispensed is represented by a uniformly distributed random variable x:
(a) Find the mean and standard deviation of x.
(b) Find the probability that x is at least 7 ounces.
(c) Find the probability that x is between 6.5 and 7.25 ounces.

Answer 1

Answer:

a) The mean is 7 ounces and the standard deviation is of 0.29 ounces.

b) 50% probability that x is at least 7 ounces.

c) 75% probability that x is between 6.5 and 7.25 ounces.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X at least x is given by the following formula.

$P(X \geq x) = 1 - \frac{x - a}{b-a}$

The probability that we find a value X between c and d is given by the following formula.

$P(c \leq X \leq d) = \frac{d - c}{b-a}$

The mean of the uniform distribution is:

$M = \frac{a + b}{2}$

The standard deviation of the uniform distribution is:

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

Anywhere between 6.5 and 7.5 ounces.

This means that $a = 6.5, b = 7.5$

(a) Find the mean and standard deviation of x.

Mean

$M = \frac{6.5 + 7.5}{2} = 7$

Standard deviation

$S = \sqrt{\frac{{7.5-6.5)^{2}}{12}} = 0.29$

The mean is 7 ounces and the standard deviation is of 0.29 ounces.

(b) Find the probability that x is at least 7 ounces.

$P(X \geq 7) = 1 - \frac{7 - 6.5}{7.5 - 6.5} = 0.5$

50% probability that x is at least 7 ounces.

(c) Find the probability that x is between 6.5 and 7.25 ounces.

$P(6.5 \leq X \leq 7.25) = \frac{7.25 - 6.5}{7.5 - 6.5} = 0.75$

75% probability that x is between 6.5 and 7.25 ounces.