18. With an X=60 and S=5, what proportion of scores lies above the following values?
a.65

b.47

c.55


To find the proportion of scores that lie above certain values, we'll use the **Z-score** formula. The Z-score tells us how many standard deviations a value is from the mean. The formula for the Z-score is:

\[
Z = \frac{X - \mu}{\sigma}
\]

Where:
- \( X \) is the value in question.
- \( \mu \) is the mean (here \( \mu = 60 \)).
- \( \sigma \) is the standard deviation (here \( \sigma = 5 \)).

Once we calculate the Z-scores, we'll use standard normal distribution tables (or a calculator) to find the proportion of scores above the given values.

Let's calculate the Z-scores for each value:

---

### a. \( X = 65 \)
\[
Z = \frac{65 - 60}{5} = \frac{5}{5} = 1
\]

The Z-score is 1. From the standard normal distribution, the proportion of scores **below** a Z-score of 1 is approximately 0.8413. So, the proportion of scores **above** this value is:
\[
1 - 0.8413 = 0.1587
\]

Thus, about **15.87%** of the scores lie above 65.

---

### b. \( X = 47 \)
\[
Z = \frac{47 - 60}{5} = \frac{-13}{5} = -2.6
\]

The Z-score is -2.6. From the standard normal distribution, the proportion of scores **below** a Z-score of -2.6 is approximately 0.0047. So, the proportion of scores **above** this value is:
\[
1 - 0.0047 = 0.9953
\]

Thus, about **99.53%** of the scores lie above 47.

---

### c. \( X = 55 \)
\[
Z = \frac{55 - 60}{5} = \frac{-5}{5} = -1
\]

The Z-score is -1. From the standard normal distribution, the proportion of scores **below** a Z-score of -1 is approximately 0.1587. So, the proportion of scores **above** this value is:
\[
1 - 0.1587 = 0.8413
\]

Thus, about **84.13%** of the scores lie above 55.

---

### Summary of Results:
a. About **15.87%** of scores lie above 65.  
b. About **99.53%** of scores lie above 47.  
c. About **84.13%** of scores lie above 55.

Would you like further details on any part of this calculation?

---

**Relative Questions:**
1. What is a Z-score and why is it useful?
2. How can the normal distribution be applied in real-world scenarios?
3. What happens if the standard deviation increases? How does that affect the proportions?
4. How do we interpret negative Z-scores in these problems?
5. How do we use Z-scores to find values between two numbers?

**Tip:** The Z-score gives a way to standardize values across different distributions, making comparisons easier even if the data comes from different sources.

18. With an X=60 and S=5, what proportion of scores lies above the following values?
a. 65
b. 47
c. 55

Discover how to calculate the proportion of scores above values like 65, 47, and 55 using Z-scores. With a mean of 60 and a standard deviation of 5, this problem walks through the Z-score formula and normal distribution tables. Ideal for students in Grades 9-12.

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