Question: 2 points Under current rules, the weight of a MLB baseball weighs between 142 and 149 grams. One…

2 points

Under current rules, the weight of a MLB baseball weighs between 142 and 149 grams. One particular company which makes baseballs produces baseballs approximately normally distributed, with a mean of 145 grams and standard deviation of 1.38 grams.

What is the probability that a ball chosen at random from this company would be too heavy (i.e., more than 149 grams) according to MLB standards?

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3 points

Under current rules, the weight of a MLB baseball weighs between 142 and 149 grams. One particular company which makes baseballs produces baseballs approximately normally distributed, with a mean of 145 grams and standard deviation of 1.38 grams.

Solution

The problem involves finding the probability that a baseball produced by the company is too heavy (i.e., more than 149 grams). Given: - Mean (\(\mu\)) = 145 grams - Standard deviation (\(\sigma\)) = 1.38 grams - Too heavy is defined as a weight greater than 149 grams We need to find the probability \(P(X > 149)\). First, calculate the z-score for 149 grams using the formula: \[ z = \frac{X - \mu}{\sigma} \] \[ z = \frac{149 - 145}{1.38} \] \[ z \approx \frac{4}{1.38} \approx 2.90 \] Next, look up the z-score of 2.90 in the standard normal distribution table or use a calculator to find the probability. The value will give us \(P(Z < 2.90)\). Since we want \(P(X > 149)\), this is equivalent to: \[ P(X > 149) = 1 - P(Z < 2.90) \] From the standard normal distribution table: \[ P(Z < 2.90) \approx 0.9981 \] Thus: \[ P(X > 149) = 1 - 0.9981 = 0.0019 \] Therefore, the probability that a baseball is more than 149 grams is approximately 0.0019.