To determine [tex]\( P(x \leq 104) \)[/tex] for a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 98 and a standard deviation ([tex]\(\sigma\)[/tex]) of 6, follow these steps:
1. Calculate the Z-score:
The Z-score formula is:
[tex]\[
Z = \frac{x - \mu}{\sigma}
\][/tex]
Where:
[tex]\[
x = 104, \quad \mu = 98, \quad \sigma = 6
\][/tex]
Plug in the values:
[tex]\[
Z = \frac{104 - 98}{6} = \frac{6}{6} = 1
\][/tex]
2. Find the cumulative probability:
After finding the Z-score, we look up this value on the standard normal distribution table or use a computational tool to find the cumulative probability up to [tex]\( Z = 1 \)[/tex].
The cumulative probability for [tex]\( Z = 1 \)[/tex] is approximately [tex]\( 0.8413 \)[/tex].
Therefore, [tex]\( P(x \leq 104) \)[/tex] for this normal distribution is:
[tex]\[ \boxed{0.84} \][/tex]
Out of the provided options:
A. 0.16
B. 0.975
C. 0.84
D. 0.025
The correct answer is [tex]\( \boxed{\text{C}} \)[/tex].
