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The probability calculation in the context of a lognormal distribution, such as the one given for the webpage viewing time problem, involves transforming our lognormal variable to a normal one. Since the lognormal distribution is defined as the exponentiation of a normal distribution (i.e., if \( T \) is lognormally distributed, \( T = e^X \), where \( X \) is normally distributed), we work with the natural logarithm to perform our calculations.

To find the probability that the page is viewed for more than 10 seconds, we calculate \( P(T > 10) \). First, we transform this into \( X \) by applying the logarithm: \( \ln(T) > \ln(10) \). This gives us \( \ln(10) = 2.302 \).

Next, we standardize this to use normal distribution tables or a calculator. The Z-score is computed as \( Z = \frac{\ln(10) - \mu}{\sigma} \), where \( \mu = 0.5 \) and \( \sigma = 1 \). Plug in the values to get \( Z > \frac{2.302 - 0.5}{1} = 1.802 \). Use standard normal distribution tables to find \( P(Z > 1.802) \). This yields the probability that a page is viewed for more than 10 seconds.

The median of a lognormal distribution is a straightforward calculation because it comes directly from the median of the normally distributed variable \( X \). The relationship is:\[\text{Median of lognormal } = e^{\mu}.\]

Because the median of \( X \sim N(\mu, \sigma^2) \) is \( \mu \), exponentiating \( \mu \) gives us the median of \( T \). In this exercise, \( \mu = 0.5 \).

Therefore, the median of the lognormal distribution, the point by which 50% of users move to another page, is calculated as \( e^{0.5} \). This results in approximately 1.6487 seconds. This means that half of the users spend this amount of time or less on the page.

The mean and standard deviation for a lognormal distribution require a bit more computation, using the parameters \( \mu \) and \( \sigma^2 \). For the mean of a lognormal distribution, the formula is:
\[\text{Mean } = e^{\mu + \frac{\sigma^2}{2}}.\]
For the given values, we have \( \mu = 0.5 \) and \( \sigma^2 = 1 \). Substitute these in:
\( e^{0.5 + 0.5} = e^1 \approx 2.7183 \).

The standard deviation requires more detailed computation with the formula:
\[\text{Standard Deviation} = \sqrt{(e^{\sigma^2} - 1) \cdot e^{2\mu + \sigma^2}}.\]

For the given parameters (i.e., \( \sigma^2 = 1 \) and \( \mu = 0.5 \)):
\( \sqrt{(e^1 - 1) \cdot e^{1.5}} \approx \sqrt{1.7183 \times 4.4817} \approx 3.313 \).

Here, you first compute \( e^{\sigma^2} - 1 = e^1 - 1 = 1.7183 \) and \( e^{1.5} = 4.4817 \), then use them to calculate the standard deviation. This information helps describe how spread out the viewing times are around the mean, within this lognormal distribution.