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We are given the random variable X that follow an exponential distribution such as: X ~ Exp 1. Find the expected value The expected value (mean) of an exponential distribution with parameter λ is given by E(X) = 1/λ. Therefore, for X ~ Exp(1/9.848) the expected value is E(X) = 1 / (1/9.848) = 9.848. 2. Find the standard deviation The standard deviation of an exponential distribution with parameter λ is given by σ = 1/λ. Therefore, for X ~ Exp(1/9.848) the standard deviation is σ = 1 / (1/9.848) = 9.848. 3. Find P(X<12) To find P(X < 12), i will use the cumulative distribution function (CDF) of the exponential distribution, which is F(x) = 1 - e^(-λx). Therefore, P(X < 12) = 1 - e^(-(1/9.848)*12) ≈ 0.783. 4. Find P(8<X<14) To find P(8 < X < 14), I will subtract the CDF values at 8 and 14.
P(8 < X < 14) = F(14) - F(8) = e^(-(1/9.848)*8) - e^(-(1/9.848)*14) ≈ 0.220. Reference Illowsky, B., Dean, S., Birmajer, D., Blount, B., Boyd, S., Einsohn, M., Helmreich, Kenyon, L., Lee, S., & Taub, J. (2022). Introductory statistics. openstax. https://openstax.org/books/statistics/pages/5-introduction
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