Question

The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.7 ounces and standard deviation 0.11 ounces. What is the probability that the average weight of a bar in a simple random sample (SRS) with three of these chocolate bars is between 7.56 and 7.84 ounces?

a) 0.1003
b) 0.1814
c) 0.8413
d) 0.3186

Answer 1

To find the probability that the average weight of a bar in a simple random sample with three bars is between 7.56 and 7.84 ounces, we need to calculate the z-scores and find the area under the normal distribution curve.

Step-by-step explanation:

To find the probability that the average weight of a bar in a simple random sample with three bars is between 7.56 and 7.84 ounces, we need to calculate the z-scores for both values and find the area between them under the standard normal distribution curve.

First, we need to calculate the z-scores:

Z1 = (7.56 - 7.7) / (0.11 / sqrt(3)) = -2.909

Z2 = (7.84 - 7.7) / (0.11 / sqrt(3)) = 1.818

Using a standard normal distribution table or a calculator, we can find that the area to the left of Z1 is 0.0019 and the area to the left of Z2 is 0.9641. Therefore, the probability that the average weight falls between 7.56 and 7.84 ounces is 0.9641 - 0.0019 = 0.9622.