Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of 1.
If P(0<z<a)=.1064 , find a.
Since the z-scores are normally distributed with a mean of 0 and a standard deviation of 1, we can find the corresponding z-values using the standard normal distribution table (also known as the Z-table) or a calculator.
The given probability is P(0 < z < a) = 0.1064.
Looking at the standard normal distribution table, we need to find the z-values that correspond to a cumulative probability of 0.1064 for P(0 < z < a).
From the table, we can find that the z-value that corresponds to a cumulative probability of 0.1064 is approximately -1.18.
Therefore, a standard deviation (z-value) that corresponds to a cumulative probability of 0.1064 is -1.18.
Thus, a = -1.18.