The amount of soda a dispensing machine pours into a 12-ounce can of soda is normally distributed with a mean of 12.03 ounces and a standard deviation of 0.02 ounce. What proportion of the soda cans contain more than the advertised 12 ounces of soda?

93.32% of the soda cans contain more than the advertised 12 ounces of soda

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

What proportion of the soda cans contain more than the advertised 12 ounces of soda?

This is 1 subtracted by the pvalue off Z when X = 12. So

has a pvalue of 0.0668

1 - 0.0668 = 0.93332

93.32% of the soda cans contain more than the advertised 12 ounces of soda

the proportion of the soda cans that contain more than the advertised 12 ounces of soda is 0.933

Step-by-step explanation:

Since the amount of soda that the dispensing machine pours into a 12-ounce can of soda is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = amount of soda poured into the soda cans

µ = mean amount

σ = standard deviation

From the information given,

µ = 12.03 ounces

σ = 0.02 ounce

The probability that the soda cans contain more than the advertised 12 ounces of soda is expressed as

P(x > 12) = 1 - P(x ≤ 12)

For x = 12,

z = (12 - 12.03)/0.02 = - 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.067

P(x > 12) = 1 - 0.067

P(x > 12) = 0.933

0 is its x and y coordinate

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