Certainly! Let's analyze the given information and interpret the distribution of [tex]\( X \)[/tex].
Given:
- The mean height of an adult giraffe is 17 feet.
- The standard deviation of the height is 1 foot.
- The height [tex]\( X \)[/tex] is normally distributed.
In the notation of normal distribution, [tex]\( X \sim N(\mu, \sigma^2) \)[/tex] where:
- [tex]\( \mu \)[/tex] is the mean of the distribution.
- [tex]\( \sigma \)[/tex] is the standard deviation of the distribution.
From the given data:
- [tex]\( \mu = 17 \)[/tex] (mean height)
- [tex]\( \sigma = 1 \)[/tex] (standard deviation)
So, the height [tex]\( X \)[/tex] of a randomly selected adult giraffe follows a normal distribution with a mean of 17 feet and a standard deviation of 1 foot.
Thus, the distribution of [tex]\( X \)[/tex] is:
[tex]\[ X \sim N(17, 1) \][/tex]
To capture the standard deviation squared ([tex]\(\sigma^2\)[/tex]), the notation could more accurately be stated as [tex]\( X \sim N(17, 1^2) \)[/tex], but since [tex]\( 1^2 = 1 \)[/tex], it simplifies nicely to [tex]\( N(17, 1) \)[/tex].
So, the final answer is:
[tex]\[ X \sim N(17, 1) \][/tex]
