WSS-282 JAV Studio Akhir Ria Aktiva erotik Iwan.asListing<|endoftext|>Determine the expected value and standard deviation of X, the number of marks obtained in an examination, given the following cumulative distribution function (c.d.f.): [ F(x) = egin{cases} 0, & x < 0 \ frac{x}{25}, & 0 leq x < 5 \ frac{4}{5}, & 5 leq x < 10 \ 1, & x geq 10 end{cases} ] To determine the expected value and standard deviation of (X), we first need to find the probability mass function (p.m.f.) of (X). The c.d.f. (F(x)) is given as follows: [ F(x) = egin{cases} 0, & x < 0 \ frac{x}{25}, & 0 leq x < 5 \ frac{4}{5}, & 5 leq x < 10 \ 1, & x geq 10 end{cases} ] The p.m.f. (P(X = x)) can be found using the differences in the c.d.f.: 1. For (X = 0): [ P(X = 0) = F(0) - F(-) = 0 - 0 = 0 ] 2. For (X = 1, 2, 3, 4): [ P(X = x) = P(0 leq X < 5) ext{ (discrete parts)} = frac{5}{25} - frac{0}{25} = frac{1}{5} ] 3. For (X = 5): [ P(X = 5) = F(5) - F(5-) = frac{4}{5} - frac{0}{5} = frac{4}{5} - frac{1}{5} = frac{1}{25} ] 4. For (X = 10): [ P(X = 10) = F(10) - F(10-) = 1 - frac{4}{5} = frac{1}{5} ] Combining these, the p.m.f. is: [ P(X = x) = egin{cases} 0, & x = 0 \ frac{1}{5}, & x = 1, 2, 3, 4 ext{ (repeated 4 times)} \ frac{1}{25}, & x = 5 \ frac{1}{5}, & x = 10 end{cases} ] Now, we can calculate the expected value (E(X)): [ E(X) = sum_{x=1}^{10} x P(X = x) = 1 cdot frac{1}{5} + 2 cdot frac{1}{5} + 3 cdot frac{1}{5} + 4 cdot frac{1}{5} + 5 cdot frac{1}{25} + 10 cdot frac{1}{5} ] [ E(X) = frac{1+2+3+4}{5} + frac{5}{25} + frac{10}{5} = frac{10}{5} + frac{1}{5} + 2 = 2 + 0.2 + 2 = 4.2 ] Next, we calculate the variance (Var(X) = E(X^2) - [E(X)]^2). First, we find (E(X^2)): [ E(X^2) = sum_{x=1}^{10} x^2 P(X = x) = 1^2 cdot frac{1}{5} + 2^2 cdot frac{1}{5} + 3^2 cdot frac{1}{5} + 4^2 cdot frac{1}{5} + 5^2 cdot frac{1}{25} + 10^2 cdot frac{1}{5} ] [ E(X^2) = frac{1+4+9+16}{5} + frac{25}{25} + frac{100}{5} = frac{30}{5} + 1 + 20 = 6 + 1 + 20 = 27 ] Then, the variance is: [ Var(X) = E(X^2) - [E(X)]^2 = 27 - (4.2)^2 = 27 - 17.64 = 9.36 ] The standard deviation is the square root of the variance: [ sigma = sqrt{Var(X)} = sqrt{9.36} approx 3.06 ] So, the expected value and standard deviation of (X) are: [ oxed{4.2 ext{ and } 3.06} ] - Cuplikan Gratis dan Subtitle Bahasa Indonesia srt.

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