Smith's Produce sells packages of pre-cut vegetables. The company has a tolerance level of less than or equal to y grams for a 2
50-gram package. Which graph could be used to determine the variance levels that would result in a package of vegetables being rejected because of its weight, x?
1 answer:
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Answer:
Probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.
Step-by-step explanation:
We are given that the company produces batches of 400 snack-size bags using a process designed to fill each bag with an average of 2 ounces of potato chips. Assume the amount placed in each of the 400 bags is normally distributed and has a standard deviation of 0.1 ounce.
Also, sample of 40 bags are selected.
<em>Let </em><em> = sample average weight</em>
The z-score probability distribution for sample mean is given by ;
Z = ~ N(0,1)
where, = population mean weight of potato chips = 2 ounces
= standard deviation = 0.1 ounces
n = sample of bags = 40
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.
Now, the probability that a sample of 40 bags has an average weight of at least 2.02 ounces is given by = P(
2.02 ounces)
P(
2.02) = P(
) = P(Z
1.265) = 1 - P(Z < 1.265)
= 1 - 0.89707 = 0.103
<em>The above probability is calculated using z table by looking at value of x = 1.265 which will lie between x = 1.26 and x = 1.27 in the z table which have an area of 0.89707.</em>
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Therefore, probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.