Mathematics, 14.04.2020 20:15 bricksaspares

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.

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Answer from: daryn1234

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

50% probability that x is at least 7 ounces.

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

Given that,

New beverage- dispensing machine, which is designed to dispense 7 ounces, in fact dispenses anywhere between 6.5 and 7.5 ounces.

We have to determine,

The amount dispensed is represented by a uniformly distributed random variable x.

According to the question,

The continuous uniform distribution or rectangular distribution is a family of symmetric probability distributions.

In a uniform distribution the probability of each event is exactly the same.

For this situation, A lower limit of the distribution denoted as a that an upper limit denoted as b that ,

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

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

The probability that 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 uniform distribution is given by,

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

And The standard deviation of uniform distribution is given by,

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

Anywhere between 6.5 and 7.5 ounces .

Then, a = 6.5 and b = 7.5

Mean and standard deviation of x is given by,

$M = \frac{6.5+7.5}{2} \\\\M = \frac{14}{2} \\\\M = 7$

The value of mean is 7.

Standard deviation is given by,

$s = \sqrt{\frac{(7.5-6.5)^{2} }{12} }\\\\s = \sqrt{\frac{(1.5)^{2}} {12}$

$s = \sqrt{ \frac{2.25}{12} }\\\\s = \sqrt{0.18} \\\\s = 0.424$

The standard deviation is of 0.29 ounces.

The probability that x is at least 7 ounces.

$P(x\geq 1) = 1 - \frac{7-6.5}{7.5-6.5} \\\\p(x\geq 1) = 1- \frac{0.5}{1.5} \\\\P(x\geq 1) = \frac{1.5-0.5}{1.5} \\\\ P(x\geq 1) = 0.5$

50% probability that x is at least 7 ounces.

The probability that x is between 6.5 and 7.25 ounces,

$P(6.5\leq X\leq 7.5) = \frac{7.25-6.5}{7.5-6.5} = \frac{0.75}{1} = 0.75$

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

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Answer from: ultimatesaiyan

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.

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

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

Answer from: Quest

function 2 has a greater rate of change than function 1

function 2 has a greater y-intercept than function 1

step-by-step explanation:

$\newline{\text{the rate of change is change per unit time}}$ $\newline{\text{for (x,y), rate of change is }\frac{y_2-y_1}{x_2-x_1} \text{ or change in y divided by change in x}}$ $\newline{\text{for function 1, the cost goes up by 75-45=30 each time every hour}}$ $\newline{\text{that means that the slope of function 1 is 30 per hour}}$ $\newline{\text{for the 2nd function, 40 is added every hour. therefor, the slope is 40 per hour}}$ $\bigskip\newline{\text{since 40 per hour> 30 per hour, function 2 has a greater rate of change than function 1}}$ $\bigskip\newline{\text{the y intercept is the y value when x is 0}}$ $\newline{\text{for function 1, the y value when x is 0 is 45}}$ $\newline{\text{for function 2, the y value when x is 0 (or the flat rate) is 50}}$ $\newline{\text{since 50> 45, function 2 has a greater y intercept than funciton 1}}$ $\bigskip\bigskip\newline{\text{the answers are: }}$ $\newline{\text{ function 2 has a greater rate of change than function 1}}$ $\newline{\text{function 2 has a greater y-intercept than function 1.}}$

Answer from: Quest

The answer is 76° i believe

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