Solution: First we need to find the mean of this distribution. The mean is. ) = e[( e 1 e 1 ) 0] = 2.

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1 Math 0A with Professor Stankova Worksheet, Discussion #; Monday, //207 GSI name: Roy Zhao Standard Deviation Example. Let f(x) = e e x for x and 0 otherwise. Find the standard deviation of this distribution. Solution: First we need to find the mean of this distribution. The mean is x(e e x )dx + 0dx = e = e(xe x e x ) = e[( e e ) 0] = 2. xe x dx To find the standard deviation, we first find the variance and then take the square root. There are two ways to do this, the latter is a bit easier (x ( 2)) 2 f(x)dx = x 2 f(x)dx µ 2 = x 2 (e e x )dx 4 = e(x 2 e x 2xe x + 2e x ) 4 = e(e + 2e + 2e ) 4 = 5 4 =. So the standard deviation is σ =. 2. Find the standard deviation of the set {, 2, }. Solution: First we find the mean, which is the average, and hence it is +2+ = 2. Then we calculate the variance which is ( 2) 2 + (2 2) 2 + ( 2) 2 = 2. For statisticians, or something called an unbiased estimator, we divide by N or 2 instead to get a variance of. The standard deviation is thus 6/ or. Problems. True FALSE The standard deviation always exists.

2 Solution: The standard deviation requires the mean to exist, and sometimes that doesn t exist. 4. True FALSE Sometimes, we take the standard deviation to be the negative square root of the variance. Solution: The standard deviation is always nonnegative. 5. TRUE False The variance is always nonnegative. Solution: The variance is (x µ) 2 f(x)dx and both (x µ) 2 0 and f(x) 0 so (x µ) 2 f(x) 0 so the integral must be nonnegative too. 6. TRUE False If the mean doesn t exist, then the standard deviation doesn t exist. Solution: The formula for the standard deviation requires the mean, so if the mean doesn t exist, then we can t talk about the standard deviation. 7. True FALSE If the mean exists, then the standard deviation exists. Solution: It is possible for the mean to exist but the standard deviation not to exist. For example, the distribution on x has the mean existing but the standard x deviation not. 8. Let f(x) be 2/x from x 2 and 0 everywhere else. Find the standard deviation of this distribution. (x 4/9) 2 f(x)dx = Thus, σ = /62 = 26/8. 2 2/x 2 dx = 2 9 x 2 = /x dx (4/9) 2 = = 62.

3 9. Let f(x) be 4/x 5 for x and 0 everywhere else. Find the standard deviation of this distribution. (x ( 4/)) 2 f(x)dx = Thus, σ = 2/9 = 2/. 4/x 4 dx = 4 x = 4. 4/x dx ( 4/) 2 = = Let f(x) be the uniform distribution on 0 x 0 and 0 everywhere else. Find the standard deviation of this distribution. Solution: Since f is the uniform distribution on [0, 0], we know that f(x) = on [0, 0] and 0 everywhere else. First we find the mean as 0 (x 5) 2 f(x)dx = Thus, σ = 25/ = 5 /. 0 0 x/0dx = x 2 / = = x 2 /0dx 5 2 = =.. Let f(x) be 2x from x 0 and 0 everywhere else. Find the standard deviation of this distribution. (x ( 2/)) 2 f(x)dx = Thus, σ = /8 = 2/6. 0 2x 2 dx = 2/x 0 = x dx ( 2/) 2 = = 8.

4 2. Let f(x) be 24/x 4 for x 2 and 0 everywhere else. Find the standard deviation of this distribution. 2 Thus, σ =. (x ) 2 f(x)dx = 24/x dx = 2x 2 2 =. x 2 f(x)dx µ 2 = 2 24/x 2 dx 2 = 2 9 =.. Let f(x) be the uniform distribution on 20 x 0 and 0 everywhere else. Find the standard deviation of this distribution. Solution: Since f is the uniform distribution on [ 20, 0], we know that f(x) = on [ 20, 0] and 0 everywhere else. First we find the mean as = 0 ( 20) x/0dx = x 2 / = 5. This could also be found by noting that the distribution is symmetric around 5 and that it is finite. (x ( 5)) 2 f(x)dx = Thus, σ = 25/ = 5 / x 2 /0dx ( 5) 2 = =. Chebyshev s Inequality Example 4. Let f(x) = e e x for x and 0 otherwise. Estimate the probability P ( 4 X 0). For what a can we say that P (X a) 0.99?

5 Solution: Since the mean is 2 and the standard deviation is, using Chebyshev s inequality, we have that P ( 4 X 0) = P (µ 2σ X µ + 2σ) = So an estimate would be. The real answer is In order for P (X a) 0.99, we set a = µ kσ and we know that P (µ kσ X µ + kσ). We want to set this lower bound to 0.99 and doing so gives k 2 k = 0. Thus, we have that P (µ 0σ X µ + 0σ) = P ( 2 X 8) = P ( 2 X) So, we have that a = 2. Problems 5. True FALSE Chebyshev s inequality can tell us what the probability actually is. Solution: Like error bounds, Chebyshev s inequality just gives us an estimate and not the actual probability. 6. True FALSE For Chebyshev s inequality, the k must be an integer. Solution: We can take k to be any positive real number. 7. True FALSE Chebyshev s inequality can help us estimate P (µ σ X µ + σ). Solution: Using Chebyshev s inequality, we get that this probability is greater than / = 0 which we knew anyway because it is a probability. 8. Let f(x) be 2/x from x 2 and 0 everywhere else. Estimate P (0/9 X 2). Solution: The mean is 4/9 and so this probability is P (4/9 4/9 X 4/9 + 4/9). Letting 4/9 = kσ = k 268, we calculate that k = Then, we have that P (0/9 X 2) /k 2 = /(64/26) = Let f(x) be 4/x 5 for x and 0 everywhere else. Estimate P (X )

6 Solution: The mean is 4/ and since f(x) = 0 for all x >, we have that P (X 2) = P ( 4/ 5/ X 4/ + 5/) /k 2. Here, we have that 5/ = kσ = k 2/ and so k = 5/ 2 and /k 2 = /(25/2) = Let f(x) be the uniform distribution on 0 x 0 and 0 everywhere else. Estimate P (2 X 8). Solution: The mean is 5 and so this probability is P (5 X 5 + ). Letting = kσ = k5 /, we calculate that k = 9 5. Then, we have that P (2 X 8) /k 2 = /(8/75) = Let f(x) be 2x from x 0 and 0 everywhere else. Estimate P ( X /). Solution: First we know that P ( X /) = P ( 2/ / X 2/ + /). Let / = kσ = k/ 8. Then k = 8 = 2 and /k 2 = k 2 /2 = / Let f(x) be 24/x 4 for x 2 and 0 everywhere else. Estimate P (X 5). Solution: The mean is and so P (X 5) = P (X + 2) = P ( 2 X + 2) because P (X ) = 0. Let 2 = kσ = k so k = 2/. Then /k 2 = /(4/) = Let f(x) be the uniform distribution on 20 x 0 and 0 everywhere else. Estimate P ( 8 X 2). Solution: The mean is 5 and so this probability is P ( 5 X 5 + ). Letting = kσ = k5 /, we calculate that k = 9 5. Then, we have that P (2 X 8) /k 2 = /(8/75) = 2. 27