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Sagot :
Let's address each part of the problem step-by-step.
Given:
- [tex]\( X \)[/tex] is normally distributed with a mean ([tex]\( \mu \)[/tex]) of 10 and a standard deviation ([tex]\( \sigma \)[/tex]) of 2.
### Part (a)
Problem:
[tex]\[ P(X > x) = 0.5 \][/tex]
Solution:
Since the total probability is 1, [tex]\( P(X \leq x) = 1 - P(X > x) = 0.5 \)[/tex]. Given that the 50th percentile (or median) of a normal distribution is the mean:
[tex]\[ x = \mu = 10 \][/tex]
### Part (b)
Problem:
[tex]\[ P(X > x) = 0.95 \][/tex]
Solution:
First, we convert this to less than form:
[tex]\[ P(X \leq x) = 1 - P(X > x) = 1 - 0.95 = 0.05 \][/tex]
We find the z-score for the 5th percentile (0.05). This z-score converts to:
[tex]\[ z \approx -1.645 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = \mu + z \cdot \sigma = 10 + (-1.645) \cdot 2 = 6.71 \][/tex]
### Part (c)
Problem:
[tex]\[ P(x < X < 10) = 0.2 \][/tex]
Solution:
We know:
[tex]\[ P(X < 10) = 0.5 \][/tex]
Thus:
[tex]\[ P(X < x) = 0.5 - 0.2 = 0.3 \][/tex]
We find the z-score for the 30th percentile (0.3). This z-score converts to:
[tex]\[ z \approx -0.524 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = \mu + z \cdot \sigma = 10 + (-0.524) \cdot 2 = 8.95 \][/tex]
### Part (d)
Problem:
[tex]\[ P(-x < X - 10 < x) = 0.95 \][/tex]
Solution:
This can be taken as:
[tex]\[ P(10 - x < X < 10 + x) = 0.95 \][/tex]
This means the probability between [tex]\( 10 - x \)[/tex] and [tex]\( 10 + x \)[/tex] is 0.95. We look for the critical value, which is symmetrically distributed around the mean. The z-scores for a 95% confidence interval are:
[tex]\[ \pm 1.96 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = z \cdot \sigma = 1.96 \cdot 2 \approx 3.92 \][/tex]
### Part (e)
Problem:
[tex]\[ P(-x < X - 10 < x) = 0.99 \][/tex]
Solution:
This can be taken as:
[tex]\[ P(10 - x < X < 10 + x) = 0.99 \][/tex]
This means the probability between [tex]\( 10 - x \)[/tex] and [tex]\( 10 + x \)[/tex] is 0.99. We look for the critical value, which is symmetrically distributed around the mean. The z-scores for a 99% confidence interval are:
[tex]\[ \approx \pm 2.576 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = z \cdot \sigma = 2.576 \cdot 2 \approx 5.15 \][/tex]
### Summary of Answers:
- Part (a): [tex]\( x = 10 \)[/tex]
- Part (b): [tex]\( x = 6.71 \)[/tex]
- Part (c): [tex]\( x = 8.95 \)[/tex]
- Part (d): [tex]\( x = 3.92 \)[/tex]
- Part (e): [tex]\( x = 5.15 \)[/tex]
Given:
- [tex]\( X \)[/tex] is normally distributed with a mean ([tex]\( \mu \)[/tex]) of 10 and a standard deviation ([tex]\( \sigma \)[/tex]) of 2.
### Part (a)
Problem:
[tex]\[ P(X > x) = 0.5 \][/tex]
Solution:
Since the total probability is 1, [tex]\( P(X \leq x) = 1 - P(X > x) = 0.5 \)[/tex]. Given that the 50th percentile (or median) of a normal distribution is the mean:
[tex]\[ x = \mu = 10 \][/tex]
### Part (b)
Problem:
[tex]\[ P(X > x) = 0.95 \][/tex]
Solution:
First, we convert this to less than form:
[tex]\[ P(X \leq x) = 1 - P(X > x) = 1 - 0.95 = 0.05 \][/tex]
We find the z-score for the 5th percentile (0.05). This z-score converts to:
[tex]\[ z \approx -1.645 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = \mu + z \cdot \sigma = 10 + (-1.645) \cdot 2 = 6.71 \][/tex]
### Part (c)
Problem:
[tex]\[ P(x < X < 10) = 0.2 \][/tex]
Solution:
We know:
[tex]\[ P(X < 10) = 0.5 \][/tex]
Thus:
[tex]\[ P(X < x) = 0.5 - 0.2 = 0.3 \][/tex]
We find the z-score for the 30th percentile (0.3). This z-score converts to:
[tex]\[ z \approx -0.524 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = \mu + z \cdot \sigma = 10 + (-0.524) \cdot 2 = 8.95 \][/tex]
### Part (d)
Problem:
[tex]\[ P(-x < X - 10 < x) = 0.95 \][/tex]
Solution:
This can be taken as:
[tex]\[ P(10 - x < X < 10 + x) = 0.95 \][/tex]
This means the probability between [tex]\( 10 - x \)[/tex] and [tex]\( 10 + x \)[/tex] is 0.95. We look for the critical value, which is symmetrically distributed around the mean. The z-scores for a 95% confidence interval are:
[tex]\[ \pm 1.96 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = z \cdot \sigma = 1.96 \cdot 2 \approx 3.92 \][/tex]
### Part (e)
Problem:
[tex]\[ P(-x < X - 10 < x) = 0.99 \][/tex]
Solution:
This can be taken as:
[tex]\[ P(10 - x < X < 10 + x) = 0.99 \][/tex]
This means the probability between [tex]\( 10 - x \)[/tex] and [tex]\( 10 + x \)[/tex] is 0.99. We look for the critical value, which is symmetrically distributed around the mean. The z-scores for a 99% confidence interval are:
[tex]\[ \approx \pm 2.576 \][/tex]
Then, we use the formula to convert the z-score to [tex]\( x \)[/tex]:
[tex]\[ x = z \cdot \sigma = 2.576 \cdot 2 \approx 5.15 \][/tex]
### Summary of Answers:
- Part (a): [tex]\( x = 10 \)[/tex]
- Part (b): [tex]\( x = 6.71 \)[/tex]
- Part (c): [tex]\( x = 8.95 \)[/tex]
- Part (d): [tex]\( x = 3.92 \)[/tex]
- Part (e): [tex]\( x = 5.15 \)[/tex]
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