To determine the probability that a score from a normal distribution with a mean ([tex]\(\mu\)[/tex]) of 60 and a standard deviation ([tex]\(\sigma\)[/tex]) of 8 will be greater than [tex]\(X = 54\)[/tex], follow these steps:
### Step 1: Calculate the Z-score
The Z-score is a way of standardizing a value from a normal distribution. It represents the number of standard deviations a data point is from the mean. The formula for the Z-score is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
For our problem:
- [tex]\(X = 54\)[/tex]
- [tex]\(\mu = 60\)[/tex]
- [tex]\(\sigma = 8\)[/tex]
Substitute these values into the formula:
[tex]\[ Z = \frac{54 - 60}{8} = \frac{-6}{8} = -0.75 \][/tex]
### Step 2: Find the Cumulative Probability for the Z-score
The cumulative distribution function (CDF) gives the probability that a random variable is less than or equal to a certain value. Using the Z-score obtained, we look up the corresponding cumulative probability for [tex]\(Z = -0.75\)[/tex].
The cumulative probability corresponding to [tex]\(Z = -0.75\)[/tex] is approximately 0.2266.
### Step 3: Calculate the Probability Greater Than X
The probability that the score is greater than [tex]\(X = 54\)[/tex] is the complement of the probability that the score is less than or equal to [tex]\(X = 54\)[/tex]. Therefore, we subtract the cumulative probability from 1:
[tex]\[ P(X > 54) = 1 - P(X \leq 54) = 1 - 0.2266 = 0.7734 \][/tex]
### Conclusion
The probability that a randomly selected score from this normal distribution is greater than 54 is approximately 0.7734.
Thus, the correct answer is:
○ A. 0.7734
