The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6.0 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.75 ounces and a standard deviation of 0.11 ounce. What percentage of the pucks produced at this factory cannot be used by the National Hockey League?
2.32
To find the percentage of pucks that cannot be used by the National Hockey League, we need to calculate the probability that a puck falls outside the weight range of 5.5 to 6.0 ounces.
Step 1: Convert the weight range to z-scores.
To do this, we'll use the formula for z-score:
z = (x - μ) / σ
For the lower weight limit of 5.5 ounces:
z1 = (5.5 - 5.75) / 0.11
For the upper weight limit of 6.0 ounces:
z2 = (6.0 - 5.75) / 0.11
Step 2: Calculate the probabilities associated with each z-score.
We'll use a standard normal distribution table or a calculator to find the probabilities.
P(z < z1) = P(z < (5.5 - 5.75) / 0.11)
P(z > z2) = P(z > (6.0 - 5.75) / 0.11)
Step 3: Calculate the percentage of pucks that cannot be used.
First, calculate the probability of a puck falling below the lower weight limit:
P(z < z1)
Next, calculate the probability of a puck falling above the upper weight limit:
P(z > z2)
Finally, subtract the two probabilities from 1 to find the percentage of pucks that cannot be used:
Percentage = (1 - P(z < z1) - P(z > z2)) * 100
Please note that z-scores are typically rounded to two decimal places when using a standard normal distribution table or calculator.
To find the percentage of pucks produced at the factory that cannot be used by the National Hockey League (NHL), we need to determine the proportion of pucks that fall outside the weight range specified by the NHL.
First, we need to find the z-scores corresponding to the lower and upper weight limits. The z-score measures the number of standard deviations a data point is away from the mean.
For the lower weight limit:
z_lower = (lower limit - mean) / standard deviation
z_lower = (5.5 - 5.75) / 0.11 = -0.25 / 0.11 ≈ -2.27
For the upper weight limit:
z_upper = (upper limit - mean) / standard deviation
z_upper = (6.0 - 5.75) / 0.11 = 0.25 / 0.11 ≈ 2.27
Next, we can use the z-scores to find the area under the normal distribution curve outside of this range using a standard normal distribution table or a statistical software.
Since we are interested in the percentage of pucks outside the range, we need to find the total area outside the range and subtract it from 1 to get the percentage.
P(z < z_lower) + P(z > z_upper) = Area outside the range
Looking up the z-scores in a standard normal distribution table, we can find the probabilities associated with those z-scores.
P(z < -2.27) ≈ 0.0116 (from the table)
P(z > 2.27) ≈ 0.0116 (from the table)
Area outside the range = P(z < -2.27) + P(z > 2.27)
≈ 0.0116 + 0.0116
≈ 0.0232
Finally, we subtract the area outside the range from 1 and convert it to a percentage to find the percentage of pucks that cannot be used by the NHL.
Percentage of pucks that cannot be used by the NHL = (1 - Area outside the range) * 100
= (1 - 0.0232) * 100
≈ 97.68%
Therefore, approximately 97.68% of the pucks produced at this factory cannot be used by the National Hockey League.