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Elenna [48]
3 years ago
14

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 an

d a standard deviation of 0.02 ounce. What proportion of the soda cans contain more than the advertised 12 ounces of soda?
Mathematics
2 answers:
borishaifa [10]3 years ago
5 0

Answer: 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

Igoryamba3 years ago
4 0

Answer:

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 \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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:

\mu = 12.03, \sigma = 0.02

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

Z = \frac{X - \mu}{\sigma}

Z = \frac{12 - 12.03}{0.02}

Z = -1.5

Z = -1.5 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

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