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2 answers:
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Answer:
0.0062 = 0.62% of the 24-ounce bags contain less than the advertised 24 ounces of chips.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean and standard deviation
, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 24.5 ounces and a standard deviation of 0.2 ounce.
This means that
What proportion of the 24-ounce bags contain less than the advertised 24 ounces of chips?
This is the p-value of Z when X = 24. So
has a p-value of 0.0062.
0.0062 = 0.62% of the 24-ounce bags contain less than the advertised 24 ounces of chips.