To calculate $P(6 \leq X \leq 14)$, we need to know the context of $X$. Typically, this would be the probability of a random variable $X$ taking values between 6 and 14 inclusive. The calculation depends on the probability distribution of $X$.

Here are the possible steps depending on the distribution:

1. If $X$ follows a continuous distribution (e.g., normal distribution):

The probability is found by integrating the probability density function (PDF) of $X$ from 6 to 14.

2. If $X$ follows a discrete distribution:

The probability is the sum of the probabilities of each value $X$ can take between 6 and 14.

Example for Normal Distribution:

If $X$ is normally distributed as $X \sim N(\mu, \sigma^2)$ (with mean $\mu$ and standard deviation $\sigma$), then the probability $P(6 \leq X \leq 14)$ is:

$P(6 \leq X \leq 14) = P\left( Z \leq \frac{14 - \mu}{\sigma} \right) - P\left( Z \leq \frac{6 - \mu}{\sigma} \right)$ where $Z$ is the standard normal variable.

Do you have information about the distribution or values for $\mu$ and $\sigma$ for this case?

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Would you like further details or have any questions? Here are 5 related questions to expand on this topic:

1. What is the probability distribution of $X$ in this problem?
2. If $X$ is normally distributed, do you have the mean ($\mu$) and standard deviation ($\sigma$) values?
3. Is $X$ a discrete or continuous random variable?
4. Do you need help calculating the cumulative probability for a specific distribution?
5. Would you like me to walk you through the normal distribution tables?

Tip: For normal distributions, always convert to the standard normal variable $Z$ using $Z = \frac{X - \mu}{\sigma}$ for easier calculations using tables or software.