abus
$a b$ and
Heights of men on a baseball team have a bell-shaped distribution with a mean of $179 \mathrm{~cm}$ and a standard deviation of $8 \mathrm{~cm}$. What is the approximate percentage of the men between the following
12 values?
$163 \mathrm{~cm}$ and $195 \mathrm{~cm}$

Step 1 ：The problem is asking for the percentage of men whose heights fall between 163 cm and 195 cm. This is a problem of normal distribution.

Step 2 ：We will use the Z-score formula to standardize the heights and then the standard normal distribution table (Z-table) to find the probabilities. The Z-score formula is: \(Z = \frac{X - \mu}{\sigma}\), where X is the value to be standardized, \(\mu\) is the mean and \(\sigma\) is the standard deviation. The Z-score represents how many standard deviations an element is from the mean.

Step 3 ：Given that the mean (\(\mu\)) is 179 cm and the standard deviation (\(\sigma\)) is 8 cm, we can calculate the Z-scores for 163 cm and 195 cm.

Step 4 ：For 163 cm, the Z-score (\(Z_1\)) is \(-2.0\) and for 195 cm, the Z-score (\(Z_2\)) is \(2.0\).

Step 5 ：We then look up these Z-scores in the Z-table to find the corresponding probabilities. The probability for \(Z_1\) is approximately 0.0228 and for \(Z_2\) is approximately 0.9772.

Step 6 ：The percentage of men between 163 cm and 195 cm is the difference between these two probabilities, which is approximately 95.45%.

Step 7 ：Final Answer: The approximate percentage of the men between 163 cm and 195 cm is \(\boxed{95.45\%}\).