The distribution of actual weights of 8-ounce chocolate bars produced by a certain machine is Normal with mean 8.1 ounces and standard deviation 0.1 ounces. Company managers do not want the weight of a chocolate bar to fall below 8 ounces, for fear that consumers will complain. Find the probability that the weight of a randomly selected candy bar is less than 8 ounces.

P(X < 8)=P((X - 8.1)/0.1 < (8 - 8.1)/0.1) = P(Z < 1)

That is, the probability that a chocolate bar falls below 8 oz is equal to the proportion of chocolate bars whose weights are *at most* +1 standard deviation above the mean.

Recall the empirical rule for normal distributions: approximately 68% of the distribution falls within 1 standard deviation of the mean, approximately 95% within 2 standard deviations, and approximately 99.7% within 3 standard deviations.

In other words,

P(-1 < Z < 1) = 0.68

P(-2 < Z < 2) = 0.95

P(-3 < Z < 3) = 0.997

which means

P(Z < -1 or Z > 1) = 1 - 0.68 = 0.32

P(Z < -2 or Z > 2) = 1 - 0.95 = 0.05

P(Z < -3 or Z > 3) = 1 - 0.997 = 0.003

The distribution is symmetric, which means

P(-1 < Z < 1) = 2 P(0 < Z < 1)

P(Z < 0) = P(Z > 0) = 0.5

The distribution is also continuous, which means

P(0 < Z < 1) = P(Z < 1) - P(Z < 0)

Putting all these facts together, we have

P(-1 < Z < 1) = 2 P (0 < Z < 1) = 2 (P(Z < 1) - P(Z < 0))

0.68 = 2 (P(Z < 1) - 0.5)

P(Z < 1) = 0.84

answer; (a) loss of judgment;

i think the answer is c) ability to survive in high salinity.

hope it