To determine [tex]\( P(x \geq 104) \)[/tex] for a normal distribution with a mean of 98 and a standard deviation of 6, we can follow these steps:
1. Understand the Problem:
- We need to find the probability that a value [tex]\( x \)[/tex] from a normal distribution is greater than or equal to 104.
- The given normal distribution has a mean ([tex]\(\mu\)[/tex]) of 98 and a standard deviation ([tex]\(\sigma\)[/tex]) of 6.
2. Standardize the Value:
- We first need to convert the value 104 to its corresponding z-score in the standard normal distribution.
- The z-score is calculated using the formula:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
- Substituting the given values:
[tex]\[
z = \frac{104 - 98}{6} = \frac{6}{6} = 1
\][/tex]
3. Find the Corresponding Probability:
- Now, we need to find the probability that [tex]\( z \)[/tex] is greater than or equal to 1 in the standard normal distribution.
- This is represented as [tex]\( P(z \geq 1) \)[/tex].
- The standard normal distribution table or calculator provides the cumulative probability up to a given z-score. For [tex]\( z = 1 \)[/tex], the cumulative probability [tex]\( P(z \leq 1) \)[/tex] is approximately 0.8413447460685429.
4. Determine the Desired Probability:
- Since we need [tex]\( P(z \geq 1) \)[/tex], we calculate it as:
[tex]\[
P(z \geq 1) = 1 - P(z \leq 1)
\][/tex]
- Substituting the cumulative probability:
[tex]\[
P(z \geq 1) = 1 - 0.8413447460685429 = 0.15865525393145707
\][/tex]
Therefore, the probability [tex]\( P(x \geq 104) \)[/tex] for a normal distribution with a mean of 98 and a standard deviation of 6 is approximately [tex]\( 0.1587 \)[/tex] or about 15.87%.
