Assume that the duration in minutes of a telephone conversation follows the exponential distribution fx-15e-x-5- x- 0- The probability that the conversation will exceed five minutes is

If f(x) is p.d.f of x, then nbsp;nbsp; P(a≤ x≤ b)=∫_a^bf(x)dx So, nbsp; nbsp; nbsp;nbsp; P(5 lt;x lt;∞ )=∫_5^∞f(x)dx=∫_5^∞15e^ x/5dx=1e ...

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Assume that the duration in minutes of a telephone conversation follows the exponential distribution $f (x) = \frac{1}{5} e^{- x / 5}, x \geq 0.$ The probability that the conversation will exceed five minutes is

\frac{1}{e}

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1 - \frac{1}{e}

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\frac{1}{e^{2}}

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1 - \frac{1}{e^{2}}

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Solution

The correct option is A $\frac{1}{e}$
if f(x) is p.d.f of x, then
$P (a \leq x \leq b) = \int_{a}^{b} f (x) d x$

So, $P (5 < x < \infty) = \int_{5}^{\infty} f (x) d x = \int_{5}^{\infty} \frac{1}{5} e^{- x / 5} d x = \frac{1}{e}$

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Assume that the duration in minutes of a telephone conversation follows the exponential distribution f(x)=15e−x/5, x≥0. The probability that the conversation will exceed five minutes is

The correct option is A 1eif f(x) is p.d.f of x, then P(a≤x≤b)=∫baf(x)dx So, P(5<x<∞)=∫∞5f(x)dx=∫∞515e−x/5dx=1e

Assume that the duration in minutes of a telephone conversation follows the exponential distribution $f (x) = \frac{1}{5} e^{- x / 5}, x \geq 0.$ The probability that the conversation will exceed five minutes is

The correct option is A $\frac{1}{e}$
if f(x) is p.d.f of x, then
$P (a \leq x \leq b) = \int_{a}^{b} f (x) d x$

So, $P (5 < x < \infty) = \int_{5}^{\infty} f (x) d x = \int_{5}^{\infty} \frac{1}{5} e^{- x / 5} d x = \frac{1}{e}$