Question

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.

Answer 1

Answer:

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!!!

Answer 2

Answer:

Probability that the bottle contains fewer than 12.38 ounces of beer is 0.00621.

Step-by-step explanation:

We are given that 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.

Let X = amount of beer poured by this filling machine

The z-score probability distribution is given by;

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

where, $\mu$ = mean amount of beer poured = 12.48 ounces

            $\sigma$ = standard deviation = 0.04 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, probability that the the bottle contains fewer than 12.38 ounces of beer is given by = P(X < 12,38 ounces)

    P(X < 12.38) = P( $\frac{ X -\mu}{\sigma}$ >  $\frac{ 12.38 - 12.48}{0.04}$ ) = P(Z < -2.50) = 1 - P(Z $\leq$ 2.50)

                                                              = 1 - 0.99379 = 0.00621

The above probability is calculated by looking at the value of x = 2.50 in the z table which has an area of 0.99379.

Therefore, probability that the bottle contains fewer than 12.38 ounces of beer is 0.00621.