The cumulative distribution function of X, the lifetime of a certain type of electronic device (measured in hours), is given by: $$F(x) = \begin{cases} 1 - 121/x^2 & \text{if } x \ge 11 \\ 0 & \text{else} \end{cases}$$ Find the probability that the lifetime is between 13.4 and 20.3 hours.
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4 and 20.3 hours. This can be written as P(13.4 < X < 20.3). The cumulative distribution function (CDF) is given by: $$F(x) = \begin{cases} 1 - 121/x^2 & \text{if } x \ge 11 \\ 0 & \text{else} \end{cases}$$ Step 2: For a continuous random variable, the probability Show more…
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The probability density function of X, the lifetime of a certain type of electronic device (measured in hours) is given by, f(x) = 10/x2 x> 10 and 0 else where Find the probability that the lifetime is between 6 and 15.5 hours_
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The probability density function of $X$, the lifetime of a certain type of electronic device (measured in hours), is given by $$ f(x)= \begin{cases}\frac{10}{x^{2}} & x>10 \\ 0 & x \leq 10\end{cases} $$ (a) Find $P\{X>20\}$. (b) What is the cumulative distribution function of $X$ ? (c) What is the probability that of 6 such types of devices at least 3 will function for at least 15 hours? What assumptions are you making?
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