f(x)=2\/x\u00b3 if x>1 and f(x)=0 elsewhere.
The diameter of a particle of contamination (in micrometers) is modeled with the probability density function
Determine P(4 <X<8) and P(X<2)
AP(4<X<8)=0.75 and P(X-2) 3\/8
P(4<X<8) 3\/64 and P(X2)-3\/4
None of the other three answers.
CP(4<X<8)=1\/2 and P(X2)=4.6875-10-2<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n We are given a probability density function f(x)f(x) that defines the probability distribution for the diameter of a particle of contamination: f(x)={2×3,if x>10,elsewheref(x) = \\begin{cases} \\frac{2}{x^3}, & \\text{if } x > 1 \\\\ 0, & \\text{elsewhere} \\end{cases}<\/p>\n\n\n\n The PDF f(x)f(x) is defined for values of x>1x > 1. This means that the random variable XX representing the particle diameter takes values greater than 1. The function 2×3\\frac{2}{x^3} describes the probability density for these values of xx.<\/p>\n\n\n\n For a probability density function, the total probability must be equal to 1. Thus, we can verify that the given PDF is valid by checking the integral over its domain: \u222b1\u221ef(x)\u2009dx=1\\int_{1}^{\\infty} f(x) \\, dx = 1<\/p>\n\n\n\n Let\u2019s calculate this integral: \u222b1\u221e2×3\u2009dx=[\u22121×2]1\u221e=0\u2212(\u22121)=1\\int_{1}^{\\infty} \\frac{2}{x^3} \\, dx = \\left[ -\\frac{1}{x^2} \\right]_{1}^{\\infty} = 0 – \\left( -1 \\right) = 1<\/p>\n\n\n\n Since this equals 1, the PDF is valid.<\/p>\n\n\n\n To find P(4<X<8)P(4 < X < 8), we need to compute the integral of the PDF from 4 to 8: P(4<X<8)=\u222b482×3\u2009dxP(4 < X < 8) = \\int_{4}^{8} \\frac{2}{x^3} \\, dx<\/p>\n\n\n\n We solve the integral: \u222b2×3\u2009dx=\u22121×2\\int \\frac{2}{x^3} \\, dx = -\\frac{1}{x^2}<\/p>\n\n\n\n So, evaluating from 4 to 8: P(4<X<8)=[\u22121×2]48=\u2212182+142=\u2212164+116=116\u2212164P(4 < X < 8) = \\left[ -\\frac{1}{x^2} \\right]_{4}^{8} = -\\frac{1}{8^2} + \\frac{1}{4^2} = -\\frac{1}{64} + \\frac{1}{16} = \\frac{1}{16} – \\frac{1}{64}<\/p>\n\n\n\n To simplify: 116=464,464\u2212164=364\\frac{1}{16} = \\frac{4}{64}, \\quad \\frac{4}{64} – \\frac{1}{64} = \\frac{3}{64}<\/p>\n\n\n\n Thus, P(4<X<8)=364P(4 < X < 8) = \\frac{3}{64}.<\/p>\n\n\n\n Now, let\u2019s calculate P(X<2)P(X < 2), which is the integral of f(x)f(x) from 1 to 2: P(X<2)=\u222b122×3\u2009dxP(X < 2) = \\int_{1}^{2} \\frac{2}{x^3} \\, dx<\/p>\n\n\n\n Using the same integral calculation: P(X<2)=[\u22121×2]12=\u2212122+112=\u221214+1=1\u221214=34P(X < 2) = \\left[ -\\frac{1}{x^2} \\right]_{1}^{2} = -\\frac{1}{2^2} + \\frac{1}{1^2} = -\\frac{1}{4} + 1 = 1 – \\frac{1}{4} = \\frac{3}{4}<\/p>\n\n\n\n Thus, P(X<2)=34P(X < 2) = \\frac{3}{4}.<\/p>\n\n\n\n Thus, the correct answer is:<\/p>\n\n\n\n C) P(4 < X < 8) = 3\/64 and P(X < 2) = 3\/4<\/strong><\/p>\n\n\n\n This explanation shows how we calculated the probabilities using the PDF and demonstrated that the given answer options match our results.<\/p>\n","protected":false},"excerpt":{"rendered":" f(x)=2\/x\u00b3 if x>1 and f(x)=0 elsewhere.The diameter of a particle of contamination (in micrometers) is modeled with the probability density functionDetermine P(4 <X<8) and P(X<2)AP(4<X<8)=0.75 and P(X-2) 3\/8P(4<X<8) 3\/64 and P(X2)-3\/4None of the other three answers.CP(4<X<8)=1\/2 and P(X2)=4.6875-10-2 The Correct Answer and Explanation is : We are given a probability density function f(x)f(x) that defines […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center 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1: Understand the probability density function (PDF)<\/h3>\n\n\n\n
Step 2: Calculate P(4<X<8)P(4 < X < 8)<\/h3>\n\n\n\n
Step 3: Calculate P(X<2)P(X < 2)<\/h3>\n\n\n\n
Step 4: Compare with the given options<\/h3>\n\n\n\n
\n