To find the probability [tex]\( P(x \leq 92) \)[/tex] given a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 98 and a standard deviation ([tex]\(\sigma\)[/tex]) of 6, we need to follow these steps:
1. Calculate the Z-Score:
The Z-score represents the number of standard deviations a data point (in this case, 92) is from the mean. The formula for calculating the Z-score is:
[tex]\[
Z = \frac{x - \mu}{\sigma}
\][/tex]
where:
- [tex]\( x \)[/tex] is the value we are interested in (92 in this case),
- [tex]\( \mu \)[/tex] is the mean (98),
- [tex]\( \sigma \)[/tex] is the standard deviation (6).
Substituting the given values:
[tex]\[
Z = \frac{92 - 98}{6} = \frac{-6}{6} = -1.0
\][/tex]
2. Find the Probability Corresponding to the Z-Score:
Using standard normal distribution tables, calculators, or software, we can find the cumulative probability associated with the Z-score of -1.0. This cumulative distribution function (CDF) value tells us the probability that a value is less than or equal to 92.
The cumulative probability for [tex]\( Z = -1.0 \)[/tex] is approximately 0.1587. Thus:
[tex]\[
P(x \leq 92) \approx 0.1587
\][/tex]
3. Determine the Correct Answer:
Comparing this probability with the multiple-choice options provided:
- A. 0.84
- B. 0.975
- C. 0.025
- D. 0.16
The closest value to 0.1587 is option D (0.16).
Therefore, the answer is:
D. 0.16
