The probability that of 10 randomly chosen bags of flour at least 2 are overweight be 0.68.
Given, that the weight of bags of all purpose flour is normally distributed with expected value 10 pounds and standard deviation 0.2 pounds.
As, we know that
P(X) = (1/s√2π)exp(-1/2((X-m)/s)^2)
where, m is the expected value
and s is the standard deviation.
Now, we have Expected value 10 pounds
and Standard Deviation 0.2 pounds
So, the probability that of 10 randomly chosen bags of flour at least 2 are overweight be
P(X=0) + P(X=1) = (1/0.2√2π)exp(-1/2((0-10)/0.2)^2) + (1/0.2√2π)exp(-1/2((1-10)/0.2)^2)
P(X=0) + P(X=1) = 0.32
So, Required Probability be 0.68.
Now, the probability that of 10 randomly chosen bags of flour at least 2 are overweight be 0.68
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