SOLUTION: A distribution of values is normal with a mean of 49.7 and a standard deviation of 25. Find P70, which is the score separating the bottom 70% from the top 30%. P70 = Ent

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Question 1181630: A distribution of values is normal with a mean of 49.7 and a standard deviation of 25.
Find P70, which is the score separating the bottom 70% from the top 30%.
P70 =
Enter your answer as a number accurate to 4 decimal places.
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Here's how to find P70:
1. **Find the z-score:** P70 represents the 70th percentile. We need to find the z-score that corresponds to a cumulative probability of 0.70. You can use a z-table or a calculator with statistical functions for this. The z-score for 0.70 is approximately 0.52.
2. **Use the z-score formula:** The z-score formula is:
z = (x - μ) / σ
Where:
* z is the z-score
* x is the value we're looking for (P70)
* μ is the mean (49.7)
* σ is the standard deviation (25)
3. **Solve for x (P70):**
0.52 = (x - 49.7) / 25
Multiply both sides by 25:
13 = x - 49.7
Add 49.7 to both sides:
x = 62.7
Therefore, P70 ≈ 62.7000
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