To determine the probability [tex]\( P(x \geq 86) \)[/tex] in a normal distribution with a mean of 98 and a standard deviation of 6, follow these steps:
1. Identify the parameters:
- Mean ([tex]\(\mu\)[/tex]): 98
- Standard deviation ([tex]\(\sigma\)[/tex]): 6
- Value we are interested in (x): 86
2. Calculate the z-score:
The z-score helps standardize the value of x within the distribution. It can be calculated using the formula:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
Plugging in the values, we get:
[tex]\[
z = \frac{86 - 98}{6} = \frac{-12}{6} = -2
\][/tex]
3. Find the cumulative probability for the calculated z-score:
The cumulative probability associated with a z-score of -2.0 represents the probability that a value is less than or equal to 86. Looking this up in the z-table (or using a computational tool), we find:
[tex]\[
P(Z \leq -2.0) \approx 0.0227501
\][/tex]
This is the cumulative distribution function (CDF) value for the z-score of -2.0.
4. Determine [tex]\(P(x \geq 86)\)[/tex]:
We are actually looking for the probability [tex]\(P(x \geq 86)\)[/tex], which is the complement of the cumulative probability:
[tex]\[
P(x \geq 86) = 1 - P(Z \leq -2.0) = 1 - 0.0227501 = 0.9772499
\][/tex]
Thus, the probability that [tex]\(P(x \geq 86)\)[/tex] is approximately 0.977.
Thus, the correct answer is:
D. 0.975
