Final answer:
To find the probability, convert the values to z-scores and use the standard normal distribution table. The probability that a randomly selected bottle will have a volume of 31.5 oz or less is approximately 25.78%. To ensure that 95% of the bottles exceed the amount labelled on the bottle, the manufacturer should print a volume of 34.82 oz.
Step-by-step explanation:
To find the probability, we can convert the given values to z-scores and use the standard normal distribution table.
a. To find the probability that a randomly selected bottle will have a volume of 31.5 oz or less, we need to find the z-score and look up its corresponding probability in the standard normal distribution table. The z-score can be calculated using the formula:
z = (x - mean) / standard deviation
Substituting the given values, we get:
z = (31.5 - 32.25) / 1.15 = -0.65
Looking up the z-score -0.65 in the standard normal distribution table, we find that the corresponding probability is approximately 0.2578 or 25.78%.
b. To find the volume that 95% of the bottles will exceed, we need to find the z-score that corresponds to a cumulative probability of 0.95. From the standard normal distribution table, we find that the z-score is approximately 1.645. Using the formula for z-score:
z = (x - mean) / standard deviation
Substituting the given values, we get:
1.645 = (x - 32.25) / 1.15
Solving for x, we get:
x = 1.645 * 1.15 + 32.25 = 34.82
Therefore, the manufacturer should print a volume of 34.82 oz on the bottle to ensure that 95% of the bottles exceed the amount labelled on the bottle.