Current learning objective: Using the Normal Distribution Question 21 Score: 0 of 4 points Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 268 feet and a standard deviation of 57 feet. Note: All calculations should be rounded to four decimal places. 1. If $$ X= $$ distance in feet for a fly ball, then $$ X \sim $$ Enter vour answer

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To express the distance of fly balls hit to the outfield as a normal distribution, we can denote it as follows:

1. The random variable $$ X $$ represents the distance in feet for a fly ball.
2. The distribution of $$ X $$ is normally distributed with a mean $$ \mu $$ and a standard deviation $$ \sigma $$.

Given:
- Mean $$ \mu = 268 $$ feet
- Standard deviation $$ \sigma = 57 $$ feet

Thus, we can write:

$$
X \sim N(\mu, \sigma^2)
$$

Where $$ N $$ denotes the normal distribution. The variance $$ \sigma^2 $$ is the square of the standard deviation.

Now, substituting the values:

$$
X \sim N(268, 57^2)
$$

Calculating $$ 57^2 $$:

$$
57^2 = 3249
$$

Therefore, the final expression for the distribution of $$ X $$ is:

$$
X \sim N(268, 3249)
$$

This indicates that the distance of fly balls hit to the outfield follows a normal distribution with a mean of 268 feet and a variance of 3249 square feet.
$$ X \sim N(268, 3249) $$

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