A landscaping firm has a profit of 5,000 on old lawn care there is a random variable X having density function
\(\displaystyle f(x) = 1/2(1-x) for -1\le x\le1\)
a) Find the variance
I think i got a) part correct. I need help with part B and c
This is what i have for part a
\(\displaystyle \mu = \int_{-1}^1 x [1/2(1 - x) ] dx \)
\(\displaystyle \mu = \frac{1}{2} \int_{-1}^1 (x - x^2) ] dx = \frac{1}{2}\left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^1\)
\(\displaystyle \mu = \frac{1}{2} (-\frac{4}{6}) = -\frac{1}{3}\) verified with wolfram
so,
\(\displaystyle $5,000( -\frac{1}{3}) = -$1667\)
\(\displaystyle \sigma^2 = E(X^2) - \mu^2\)
\(\displaystyle E(X^2) = \int_{-1}^1 x^2 [1/2(1 - x) ] dx \)
\(\displaystyle E(X^2) = \frac{1}{2} \int_{-1}^1 (x^2 - x^3) ] dx = \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{-1}^1 \)
\(\displaystyle E(X^2) = \frac{1}{3} \) verified with wolfram
The answer to Part a) \(\displaystyle = E(X^2) = \frac{1}{3} ? \)
Moving onto part b)
summary so far:
\(\displaystyle \mu = -\frac{1}{3} \) and \(\displaystyle E(X^2) = \frac{1}{3}\)
Solving for sigma
\(\displaystyle \sigma^2 = E(X^2) - \mu^2\)
\(\displaystyle \sigma^2 = \frac{1}{3} + (\frac{1}{3})^2 = \frac{4}{9}\)
Therefore,
\(\displaystyle \sigma = \sqrt\frac{4}{9} = \frac{2}{3}\)
k= 2 given
\(\displaystyle k\sigma = 2(\frac{2}{3}) = \frac{4}{3}\)
Chebyshev's Inequality: \(\displaystyle PR( |X - \mu | \ge k\sigma) \le \frac{1}{k^2}\)
\(\displaystyle PR( |X + \frac{1}{3}| \ge \frac{4}{3})\)
moving the \(\displaystyle \frac{1}{3}\) to the other side we now have:
\(\displaystyle ( \frac{1}{3} - \frac{4}{3}) = \frac{3}{3} = 1\)
\(\displaystyle PR( X \ge 1)\)
Now, i dont know if this is right up to this point. Since the limits seem off.. X = 1 to 1 is not correct.
Can someone see if i am right in my approach and what to do next from here?
\(\displaystyle f(x) = 1/2(1-x) for -1\le x\le1\)
a) Find the variance
I think i got a) part correct. I need help with part B and c
This is what i have for part a
\(\displaystyle \mu = \int_{-1}^1 x [1/2(1 - x) ] dx \)
\(\displaystyle \mu = \frac{1}{2} \int_{-1}^1 (x - x^2) ] dx = \frac{1}{2}\left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^1\)
\(\displaystyle \mu = \frac{1}{2} (-\frac{4}{6}) = -\frac{1}{3}\) verified with wolfram
so,
\(\displaystyle $5,000( -\frac{1}{3}) = -$1667\)
\(\displaystyle \sigma^2 = E(X^2) - \mu^2\)
\(\displaystyle E(X^2) = \int_{-1}^1 x^2 [1/2(1 - x) ] dx \)
\(\displaystyle E(X^2) = \frac{1}{2} \int_{-1}^1 (x^2 - x^3) ] dx = \frac{1}{2}\left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{-1}^1 \)
\(\displaystyle E(X^2) = \frac{1}{3} \) verified with wolfram
The answer to Part a) \(\displaystyle = E(X^2) = \frac{1}{3} ? \)
Moving onto part b)
summary so far:
\(\displaystyle \mu = -\frac{1}{3} \) and \(\displaystyle E(X^2) = \frac{1}{3}\)
Solving for sigma
\(\displaystyle \sigma^2 = E(X^2) - \mu^2\)
\(\displaystyle \sigma^2 = \frac{1}{3} + (\frac{1}{3})^2 = \frac{4}{9}\)
Therefore,
\(\displaystyle \sigma = \sqrt\frac{4}{9} = \frac{2}{3}\)
k= 2 given
\(\displaystyle k\sigma = 2(\frac{2}{3}) = \frac{4}{3}\)
Chebyshev's Inequality: \(\displaystyle PR( |X - \mu | \ge k\sigma) \le \frac{1}{k^2}\)
\(\displaystyle PR( |X + \frac{1}{3}| \ge \frac{4}{3})\)
moving the \(\displaystyle \frac{1}{3}\) to the other side we now have:
\(\displaystyle ( \frac{1}{3} - \frac{4}{3}) = \frac{3}{3} = 1\)
\(\displaystyle PR( X \ge 1)\)
Now, i dont know if this is right up to this point. Since the limits seem off.. X = 1 to 1 is not correct.
Can someone see if i am right in my approach and what to do next from here?
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