To solve the problem of finding [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 use the cumulative distribution function (CDF) of the normal distribution. Here are the steps:
1. Calculate the z-score:
The z-score tells us how many standard deviations away 92 is from the mean of the distribution.
The formula for the z-score is:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
Here, [tex]\(x = 92\)[/tex], [tex]\(\mu = 98\)[/tex], and [tex]\(\sigma = 6\)[/tex].
Plug in the values:
[tex]\[
z = \frac{92 - 98}{6} = \frac{-6}{6} = -1.0
\][/tex]
2. Find the probability corresponding to the z-score:
Next, we need to find the probability that the random variable [tex]\(x\)[/tex] is less than or equal to 92. This is equivalent to finding the cumulative probability up to the z-score of -1.0 in the standard normal distribution.
3. Consult the standard normal distribution table or CDF:
Using the cumulative distribution function for the standard normal distribution, we find that the probability corresponding to a z-score of -1.0 is approximately 0.1587.
4. Interpret the result:
So, [tex]\(P(x \leq 92)\)[/tex] is approximately 0.1587.
5. Match the probability to the closest given choice:
From the given options:
- A. 0.16
- B. 0.025
- C. 0.975
- D. 0.84
The probability 0.1587 is closest to option A, which is 0.16.
Therefore, [tex]\(P(x \leq 92)\)[/tex] is approximately 0.16, making the correct answer:
[tex]\[
\text{A. 0.16}
\][/tex]
