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Sagot :
Sure, let's solve the problem step by step.
### Part (a)
Given that [tex]\( X \)[/tex] is uniformly distributed over the interval [tex]\([-1, 1]\)[/tex], we need to determine the mean, variance, and standard deviation.
#### Mean
The mean of a continuous uniform distribution over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Mean} = \frac{a + b}{2} \][/tex]
For our specific case, [tex]\( a = -1 \)[/tex] and [tex]\( b = 1 \)[/tex], so:
[tex]\[ \text{Mean} = \frac{-1 + 1}{2} = \frac{0}{2} = 0.0 \][/tex]
#### Variance
The variance of a continuous uniform distribution over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Variance} = \frac{(b - a)^2}{12} \][/tex]
For [tex]\( a = -1 \)[/tex] and [tex]\( b = 1 \)[/tex]:
[tex]\[ \text{Variance} = \frac{(1 - (-1))^2}{12} = \frac{(1 + 1)^2}{12} = \frac{2^2}{12} = \frac{4}{12} = \frac{1}{3} = 0.333 \][/tex]
#### Standard Deviation
The standard deviation is the square root of the variance:
[tex]\[ \text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{0.333} \approx 0.577 \][/tex]
So, for part (a):
Mean [tex]\( = 0.0 \)[/tex]
Variance [tex]\( = 0.333 \)[/tex]
Standard Deviation [tex]\( = 0.577 \)[/tex]
### Part (b)
We need to find the value for [tex]\( x \)[/tex] such that [tex]\( P(-x < X < x) = 0.90 \)[/tex].
For a continuous uniform distribution over the interval [tex]\([-1, 1]\)[/tex], the cumulative distribution function (CDF) is:
[tex]\[ F(c) = \frac{c - a}{b - a} \][/tex]
We need to find [tex]\( x \)[/tex] such that:
[tex]\[ P(-x < X < x) = 0.90 \][/tex]
Since the distribution is symmetric around zero, we have:
[tex]\[ P(-x < X < x) = F(x) - F(-x) \][/tex]
Using the properties of the CDF for a symmetric uniform distribution:
[tex]\[ F(x) = 0.95 \][/tex]
So the value of [tex]\( x \)[/tex] that satisfies this condition is:
[tex]\[ x = -1 + 0.95 \cdot (1 - (-1)) = -1 + 0.95 \cdot 2 = -1 + 1.9 = 0.9 \][/tex]
Therefore:
The value of [tex]\( x \)[/tex] such that [tex]\( P(-x < X < x) = 0.90 \)[/tex] is [tex]\( x = 0.9 \)[/tex].
So for part (b), [tex]\( x = 0.9 \)[/tex].
### Part (a)
Given that [tex]\( X \)[/tex] is uniformly distributed over the interval [tex]\([-1, 1]\)[/tex], we need to determine the mean, variance, and standard deviation.
#### Mean
The mean of a continuous uniform distribution over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Mean} = \frac{a + b}{2} \][/tex]
For our specific case, [tex]\( a = -1 \)[/tex] and [tex]\( b = 1 \)[/tex], so:
[tex]\[ \text{Mean} = \frac{-1 + 1}{2} = \frac{0}{2} = 0.0 \][/tex]
#### Variance
The variance of a continuous uniform distribution over the interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \text{Variance} = \frac{(b - a)^2}{12} \][/tex]
For [tex]\( a = -1 \)[/tex] and [tex]\( b = 1 \)[/tex]:
[tex]\[ \text{Variance} = \frac{(1 - (-1))^2}{12} = \frac{(1 + 1)^2}{12} = \frac{2^2}{12} = \frac{4}{12} = \frac{1}{3} = 0.333 \][/tex]
#### Standard Deviation
The standard deviation is the square root of the variance:
[tex]\[ \text{Standard Deviation} = \sqrt{\text{Variance}} = \sqrt{0.333} \approx 0.577 \][/tex]
So, for part (a):
Mean [tex]\( = 0.0 \)[/tex]
Variance [tex]\( = 0.333 \)[/tex]
Standard Deviation [tex]\( = 0.577 \)[/tex]
### Part (b)
We need to find the value for [tex]\( x \)[/tex] such that [tex]\( P(-x < X < x) = 0.90 \)[/tex].
For a continuous uniform distribution over the interval [tex]\([-1, 1]\)[/tex], the cumulative distribution function (CDF) is:
[tex]\[ F(c) = \frac{c - a}{b - a} \][/tex]
We need to find [tex]\( x \)[/tex] such that:
[tex]\[ P(-x < X < x) = 0.90 \][/tex]
Since the distribution is symmetric around zero, we have:
[tex]\[ P(-x < X < x) = F(x) - F(-x) \][/tex]
Using the properties of the CDF for a symmetric uniform distribution:
[tex]\[ F(x) = 0.95 \][/tex]
So the value of [tex]\( x \)[/tex] that satisfies this condition is:
[tex]\[ x = -1 + 0.95 \cdot (1 - (-1)) = -1 + 0.95 \cdot 2 = -1 + 1.9 = 0.9 \][/tex]
Therefore:
The value of [tex]\( x \)[/tex] such that [tex]\( P(-x < X < x) = 0.90 \)[/tex] is [tex]\( x = 0.9 \)[/tex].
So for part (b), [tex]\( x = 0.9 \)[/tex].
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