Answer:
The error in this problem is that the Standard Normal Table is being used to determine probabilities related to the distribution of student ages, but the distribution of student ages is not necessarily a standard normal (z) distribution.
In the problem, you are given the mean age of students at a high school (16.5) and the standard deviation (0.7). This suggests that the distribution of student ages is likely to be normal, but it may not necessarily have a mean of 0 and a standard deviation of 1, which is the standard normal distribution.
To calculate the probability of selecting a student whose age is more than 17.5 years, you would typically need to use the z-score formula to standardize the age value and then use a standard normal table. The z-score formula is:
\[z = \frac{x - \mu}{\sigma}\]
Where:
- \(x\) is the value you want to standardize (in this case, 17.5).
- \(\mu\) is the mean age of the population (16.5).
- \(\sigma\) is the standard deviation of the population (0.7).
So, in this case, the z-score would be:
\[z = \frac{17.5 - 16.5}{0.7} = \frac{1}{0.7} \approx 1.43\]
Now, you can use a standard normal table to find the probability associated with a z-score of 1.43. However, since the problem doesn't specify the exact distribution of student ages, it's important to recognize that using the standard normal table in this context may not be entirely accurate. You would need more information about the distribution of student ages to make a more precise calculation.
So, the error in the problem is that it assumes a standard normal distribution for student ages without providing sufficient information to justify this assumption.
Explanation: