A regulation hockey puck must weigh between 5.5 and 6 ounces. In an alternative manufacturing process the mean weight of pucks produced is 5.75 ounce. The weights of pucks have a normal distribution whose standard deviation can be decreased by increasingly stringent (and expensive) controls on the manufacturing process. Find the maximum allowable standard deviation so that at most 0.005 of all pucks will fail to meet the weight standard. (Hint: The distribution is symmetric and is centered at the middle of the interval of acceptable weights.)
To find the maximum allowable standard deviation, we need to determine the corresponding z-score for the given probability and then find the standard deviation value associated with that z-score.
1. Calculate the z-score:
The distribution is symmetric and centered at the middle of the acceptable weight interval, which is the mean weight of 5.75 ounces.
Using a Z-table or a calculator, find the z-score that corresponds to a cumulative probability of 1 - 0.005 = 0.995 (since we want at most 0.005 pucks to fail to meet the weight standard). The z-score that corresponds to a cumulative probability of 0.995 is approximately 2.575.
2. Calculate the maximum allowable standard deviation:
The z-score formula is given by:
z = (X - μ) / σ
Rearranging the formula to solve for σ gives:
σ = (X - μ) / z
Substituting the values into the formula:
σ = (6 - 5.75) / 2.575 ≈ 0.0971
Therefore, the maximum allowable standard deviation is approximately 0.0971 ounces.