To determine the probability [tex]\( P(x \leq 93) \)[/tex] for a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 79 and a standard deviation ([tex]\(\sigma\)[/tex]) of 7, we can follow these steps:
1. Understand the Problem:
- Mean ([tex]\(\mu\)[/tex]): 79
- Standard Deviation ([tex]\(\sigma\)[/tex]): 7
- Value ([tex]\(x\)[/tex]): 93
2. Calculate the z-score:
The z-score allows us to standardize our normal distribution, converting it into the standard normal distribution ([tex]\(N(0, 1)\)[/tex]). The z-score is calculated using the formula:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
Plugging in the values:
[tex]\[
z = \frac{93 - 79}{7} = \frac{14}{7} = 2
\][/tex]
3. Find the probability associated with the z-score:
Using the cumulative distribution function (CDF) for the standard normal distribution, we can find the probability that a value is less than or equal to our calculated z-score. In this case:
[tex]\[
P(z \leq 2)
\][/tex]
Looking up the value in the standard normal distribution table or using a calculator, we find:
[tex]\[
P(z \leq 2) \approx 0.9772
\][/tex]
4. Determine the probability:
Therefore, the probability that [tex]\( x \leq 93 \)[/tex] is approximately 0.9772.
Given the choices:
A. 0.84
B. 0.16
C. 0.975
D. 0.025
The correct answer, closest to our calculated probability, is:
C. 0.975
