the pdf of a continuous random variable x is given by f(x)="\\int c\/" square root x where 0 <x<4 find the constant c and find the cumulative distribution function of x then compute P(x>1)
On of the properties of pdf is: "\\int_{-\\infty}^{+\\infty} f(x)dx = 1", where f(x) is the pdf.
In the given case: "\\int_{-\\infty}^{+\\infty} {\\frac c {\\sqrt x}}dx = 1\\to\\int_0^4 {\\frac c {\\sqrt x}}dx = 1\\to 2c\\sqrt{4}-2c\\sqrt{0}=1\\to"
"\\to 4c=1\\to c = 0.25"
The cdf is equal to 0 when x≤0, equal to 1 when x≥4, and equal to "\\int_0^x {\\frac 1 {4\\sqrt t}}dt" when x is 0<x<4
"\\int_0^x {\\frac 1 {4\\sqrt t}}dt =0.25\\int_0^x {\\frac 1 {\\sqrt t}}dt=0.25(2\\sqrt{x}-2\\sqrt{0})={\\frac {\\sqrt{x}} 2}"
"P(x>1) = 1-P(x\u22641)=1-0.5=0.5"
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