Suppose a brewery has a filling machine that fills 12 ounce bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean of 12.48 ounces and a standard deviation of 0.04 ounce. Find the probability that the bottle contains fewer than 12.38 ounces of beer.

The probability that the bottle contains fewer than 12.38 ounces of beer

P(x < 12.38) = 0.00621

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 12.48 ounces

Standard deviation = σ = 0.04 ounces

The probability that the bottle contains fewer than 12.38 ounces of beer

P(x < 12.38)

To solve this, we need to first normalize/standardize 12.38

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (12.38 - 12.48)/0.04 = -2.50

To determine the probability that the bottle contains fewer than 12.38 ounces of beer

P(x < 12.38) = P(z < -2.50)

We'll use data from the normal probability table for these probabilities

P(x < 12.38) = P(z < -2.50) = 0.00621

Hope this Helps!!!

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.