To determine the probability [tex]\( P(x \geq 92) \)[/tex] for a normal distribution with a mean [tex]\(\mu = 98\)[/tex] and a standard deviation [tex]\(\sigma = 6\)[/tex], follow these steps:
1. Calculate the z-score for [tex]\( x = 92 \)[/tex]:
The z-score formula is given by:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
Substitute [tex]\( x = 92 \)[/tex], [tex]\( \mu = 98 \)[/tex], and [tex]\( \sigma = 6 \)[/tex] into the formula:
[tex]\[
z = \frac{92 - 98}{6} = \frac{-6}{6} = -1.0
\][/tex]
2. Find the cumulative probability for the calculated z-score:
The cumulative probability for a z-score can be obtained from standard normal distribution tables or using statistical software. For [tex]\( z = -1.0 \)[/tex], the cumulative probability [tex]\( P(Z < -1.0) \)[/tex] is approximately:
[tex]\[
P(Z < -1.0) \approx 0.1587
\][/tex]
This is the probability that a randomly selected value from this distribution is less than 92.
3. Determine the probability [tex]\( P(x \geq 92) \)[/tex]:
To find the probability that [tex]\( x \)[/tex] is greater than or equal to 92, use the complement rule:
[tex]\[
P(x \geq 92) = 1 - P(x < 92)
\][/tex]
Substitute the cumulative probability found in the previous step:
[tex]\[
P(x \geq 92) = 1 - 0.1587 \approx 0.8413
\][/tex]
Therefore, the probability [tex]\( P(x \geq 92) \)[/tex] is approximately 0.8413.
Hence, the correct answer is:
C. 0.84
