Final answer:
The interval containing the mid-88% of the values in a normal distribution with a mean of 90 and a standard deviation of 5 is (82.25, 97.75).
Step-by-step explanation:
To find the interval containing the mid-88% of values in a normal distribution with a mean of 90 and a standard deviation of 5, we need to calculate the z-scores that correspond to the lower 6% (since 100%-88% = 12%, and we are leaving 6% in each tail) and the upper 94% (100%-6%) of the distribution. Using standard normal distribution tables or a calculator, we find that the z-scores approximately correspond to -1.55 (for the lower 6%) and +1.55 (for the upper 94%). To convert these z-scores to actual values on the distribution, we use the formula X = μ + Zσ.
For the lower limit, the calculation is 90 + (-1.55)(5), which equals 82.25. For the upper limit, the calculation is 90 + (1.55)(5), which equals 97.75. Therefore, the interval containing the mid-88% of values is (82.25, 97.75).