The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.8 ounces and standard deviation 0.15 ounces.
(a) What is the probability that the average weight of a bar in a Simple Random Sample (SRS) with 3 of these chocolate bars is between 7.62 and 8 ounces?
ANSWER:
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(b) For a SRS of 3 of these chocolate bars, what is the level L
such that there is a 1% chance that the average weight is less than L
?
ANSWER:
To solve this problem, we need to use the properties of the normal distribution.
(a) To find the probability that the average weight of a sample of 3 chocolate bars is between 7.62 and 8 ounces, we first need to standardize the values using z-scores.
The z-score formula is z = (x - mean) / standard deviation.
For the lower bound, the z-score is (7.62 - 7.8) / 0.15 = -1.2.
For the upper bound, the z-score is (8 - 7.8) / 0.15 = 1.33.
We then look up the corresponding probabilities in the standard normal distribution table.
Using the table or a calculator, we find that the probability of a z-score less than -1.2 is approximately 0.1151, and the probability of a z-score less than 1.33 is approximately 0.9088.
To find the probability of the average weight being between 7.62 and 8 ounces, we subtract the lower probability from the upper probability:
0.9088 - 0.1151 = 0.7937
So the probability that the average weight of a sample of 3 chocolate bars is between 7.62 and 8 ounces is approximately 0.7937.
(b) To find the level L such that there is a 1% chance that the average weight is less than L, we need to find the z-score that corresponds to a cumulative probability of 0.01.
Using the standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative probability of 0.01 is approximately -2.33.
We can use the z-score formula to find the corresponding average weight:
-2.33 = (L - 7.8) / 0.15
Solving for L, we get:
L - 7.8 = -2.33 * 0.15
L - 7.8 = -0.3495
L = 7.4505
So the level L is approximately 7.4505 ounces.