To determine the probability [tex]\( P(x \geq 92) \)[/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 for [tex]\( x = 92 \)[/tex]:
The z-score formula is:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
Substituting the given values:
[tex]\[
z = \frac{92 - 98}{6}
\][/tex]
[tex]\[
z = \frac{-6}{6}
\][/tex]
[tex]\[
z = -1
\][/tex]
2. Find the cumulative probability [tex]\( P(Z < -1) \)[/tex]:
The cumulative distribution function (CDF) of a standard normal distribution provides the probability that a standard normal variable [tex]\( Z \)[/tex] is less than a specified value.
For [tex]\( z = -1 \)[/tex]:
[tex]\[
P(Z < -1) \approx 0.1587
\][/tex]
3. Determine [tex]\( P(x \geq 92) \)[/tex]:
To find [tex]\( P(x \geq 92) \)[/tex], we need to calculate the probability that [tex]\( X \)[/tex] is greater than or equal to 92. This is the complement of [tex]\( P(X < 92) \)[/tex]:
[tex]\[
P(X \geq 92) = 1 - P(X < 92)
\][/tex]
From the previous step, we know:
[tex]\[
P(X < 92) \approx 0.1587
\][/tex]
Therefore:
[tex]\[
P(X \geq 92) = 1 - 0.1587 = 0.8413
\][/tex]
So the probability [tex]\( P(x \geq 92) \)[/tex] is approximately 0.8413, which corresponds to option C (rounded to two decimal places).
Answer: C. 0.84
