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The standard normal distribution is a key concept in statistics. It is a type of normal distribution that has a mean of 0 and a standard deviation of 1.

It is often denoted as \(Z\). The importance of the standard normal distribution is that it allows us to compare different normal distributions by converting them into a common frame.

This is done through a process called standardization, which we will discuss next.

To standardize a value from a normal distribution, we calculate its z-score. The z-score tells us how many standard deviations a particular value is away from the mean.

The formula for calculating a z-score is:\[z = \frac{X - \mu}{\sigma} \]

Where:\(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

For instance, let's use the values from the exercise:

• Mean, \(\mu = 50\)
• Standard Deviation, \(\sigma = 7\)

To find the z-score for \(X = 56\), we substitute into the formula: \[z_{56} = \frac{56 - 50}{7} = \frac{6}{7} \approx 0.86 \]

And similarly for \(X = 68\): \[z_{68} = \frac{68 - 50}{7} = \frac{18}{7} \approx 2.57 \] Now we have converted our values into the standard normal distribution.

Once we have the z-scores, we can compute the probabilities associated with them using the standard normal distribution. This involves finding the area under the curve of the standard normal distribution up to the respective z-scores.

We use standard normal distribution tables, or Z-tables, to find these areas. In the given exercise, for \(z = 0.86\), we found:\[P(Z < 0.86) \approx 0.8051 \]

And for \(z = 2.57\): \[P(Z < 2.57) \approx 0.9949 \] To find the probability of \(56 < X < 68\), we subtract the probability at \(z = 0.86\) from the probability at \(z = 2.57\): \[P(56 < X < 68) = P(Z < 2.57) - P(Z < 0.86) \approx 0.9949 - 0.8051 = 0.1898 \] This value represents the probability that our initial random variable \(X\) falls between 56 and 68.