Respuesta :
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.
Let [tex]\bar X[/tex] = sample average weight
The z-score probability distribution for sample mean is given by ;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean weight of potato chips = 2 ounces
[tex]\sigma[/tex] = 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([tex]\bar X[/tex] [tex]\geq[/tex] 2.02 ounces)
P([tex]\bar X[/tex] [tex]\geq[/tex] 2.02) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\geq[/tex] [tex]\frac{2.02-2}{\frac{0.1}{\sqrt{40} } }[/tex] ) = P(Z [tex]\geq[/tex] 1.265) = 1 - P(Z < 1.265)
= 1 - 0.89707 = 0.103
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.
Therefore, probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.
