To solve this, we need to calculate the proportion of baseballs that are too heavy, i.e., those that weigh more than 5.25 ounces. Since the weights are normally distributed, we can use the z-score formula to find the probability.

Step 1: Calculate the z-score

The z-score formula is:

$z = \frac{x - \mu}{\sigma}$

Where:

• $x$ is the value we are interested in (5.25 ounces),
• $\mu$ is the mean (5.11 ounces),
• $\sigma$ is the standard deviation (0.062 ounces).

Substituting the given values:

$z = \frac{5.25 - 5.11}{0.062} = \frac{0.14}{0.062} \approx 2.26$

Step 2: Find the probability

Next, we look up the z-score of 2.26 in the standard normal distribution table or use a calculator to find the probability corresponding to this z-score.

A z-score of 2.26 corresponds to a cumulative probability of about 0.9881. This means that 98.81% of the baseballs have a weight less than 5.25 ounces.

Step 3: Calculate the proportion of baseballs that are too heavy

Since the cumulative probability represents the proportion of baseballs that weigh less than 5.25 ounces, the proportion that are too heavy is the complement of this value:

$P(\text{too heavy}) = 1 - P(Z \leq 2.26) = 1 - 0.9881 = 0.0119$

Final Answer:

The proportion of baseballs that are too heavy for use by Major League Baseball is approximately 0.0119, or 1.19%.

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Would you like more details on the calculation or have any other questions?

Here are some related questions to consider:

1. How would the proportion change if the mean weight of the baseballs were higher?
2. What would happen if the standard deviation were larger or smaller?
3. How can we calculate the proportion of baseballs that are too light?
4. How can we find the z-score for a baseball weighing exactly 5 ounces?
5. What is the significance of the standard deviation in this context?

Tip: Always remember that the z-score represents how many standard deviations a value is away from the mean, and it can be used to calculate probabilities using normal distribution.