How is the median of an exponential distribution expressed?

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How is the median of an exponential distribution expressed?

The median of an exponential distribution can be derived from its probability density function. The exponential distribution is defined by its rate parameter, commonly denoted as θ (which is the mean). To find the median, we need to determine the value m such that the cumulative distribution function (CDF) equals 0.5. The CDF for an exponential distribution is given by: \[ F(x) = 1 - e^{-x/\theta} \] Setting the CDF equal to 0.5, we have: \[ 1 - e^{-m/\theta} = 0.5 \] Rearranging this gives: \[ e^{-m/\theta} = 0.5 \] Taking the natural logarithm of both sides results in: \[ -\frac{m}{\theta} = \ln(0.5) \] From this, we can solve for m: \[ m = -\theta \ln(0.5) \] Since \(\ln(0.5)\) is equivalent to \(-\ln(2)\), it can be expressed as: \[ m = \theta \ln(2) \] This matches the option stating that the median is expressed as \( m =

The median of an exponential distribution can be derived from its probability density function. The exponential distribution is defined by its rate parameter, commonly denoted as θ (which is the mean).

To find the median, we need to determine the value m such that the cumulative distribution function (CDF) equals 0.5. The CDF for an exponential distribution is given by:

[ F(x) = 1 - e^{-x/\theta} ]

Setting the CDF equal to 0.5, we have:

[ 1 - e^{-m/\theta} = 0.5 ]

Rearranging this gives:

[ e^{-m/\theta} = 0.5 ]

Taking the natural logarithm of both sides results in:

[ -\frac{m}{\theta} = \ln(0.5) ]

From this, we can solve for m:

[ m = -\theta \ln(0.5) ]

Since (\ln(0.5)) is equivalent to (-\ln(2)), it can be expressed as:

[ m = \theta \ln(2) ]

This matches the option stating that the median is expressed as ( m =

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