The standard normal distribution is defined by its mean \( \mu = 0 \) and standard deviation \( \sigma = 1 \). We calculated the probability that a sample mean falls within the range \([-1, 1]\):
\[
P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.0) - \Phi(-1.0) = 0.6827
\]
This indicates that approximately \( 68.27\% \) of the data falls within one standard deviation of the mean in a standard normal distribution.
For a nonstandard normal distribution, we considered a mean \( \mu = 5 \) and standard deviation \( \sigma = 2 \). We calculated the probability that a sample mean falls within the range \([4, 6]\):
\[
P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.5) - \Phi(-0.5) = 0.3829
\]
This indicates that approximately \( 38.29\% \) of the data falls within this range for the nonstandard normal distribution.
The key difference between the two distributions is that the standard normal distribution has fixed parameters \( \mu = 0 \) and \( \sigma = 1 \), while the nonstandard normal distribution can have any mean and standard deviation. Therefore, the correct answer to the question is:
The standard normal distribution has a mean of \( 0 \) and a standard deviation of \( 1 \), while a nonstandard normal distribution has a different value for one or both of those parameters.