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12. (ST5) The weight of a bag of chips isapproximately normally distributed with amean of 10.72 oz. and a standard deviationof 0.14. What is the probability that arandomly selected bag of chips is less thanthe advertised weight of 10.5 oz.?

Sagot :

μ = mean = 10.72 oz

σ =standard deviation = 0.14

x = 10.5

First, find the z-score:

z = x - μ / σ

z = (10.5 - 10.72) / 0.14 = -1.57143

Then find the value on the z-score table:

P (x < 10.5) = 0.058042

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