Math  /  Data & Statistics

QuestionFor the probability density function f(x)=f(x)= 65x2\frac{6}{5} x^{-2} for 1x61 \leq x \leq 6, and f(x)=0f(x)=0 elsewhere Find P(5<X<6)P(5<X<6) (Write in the form of an integer)

Studdy Solution
Calculate the integral:
5665x2dx=[65x1]56 \int_{5}^{6} \frac{6}{5} x^{-2} \, dx = \left[ -\frac{6}{5} x^{-1} \right]_{5}^{6}
=65(16)+65(15) = -\frac{6}{5} \left( \frac{1}{6} \right) + \frac{6}{5} \left( \frac{1}{5} \right)
=15+625 = -\frac{1}{5} + \frac{6}{25}
=525+625 = -\frac{5}{25} + \frac{6}{25}
=125 = \frac{1}{25}
Convert the probability to an integer by multiplying by 100 (to express it as a percentage):
125×100=4 \frac{1}{25} \times 100 = 4
The probability P(5<X<6) P(5 < X < 6) is:
4 \boxed{4}

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