Answered

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A normal distribution has a mean of [tex] \mu = 60 [/tex] with [tex] \sigma = 8 [/tex]. If one score is randomly selected from this distribution, what is the probability that the score will be greater than [tex] X = 54 [/tex]?

A. 0.7734
B. 0.2266
C. 0.2734
D. 0.3085

Sagot :

GinnyAnswer
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
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