The population is the 10000 children, so the mean and standard deviation is 41.26 and 2.24 (there is no approximation involved).

The distribution of the sample mean of a sample of 100 is approximately normal with mean 41.26 and standard deviation 0.224 (the population SD divided by the square root of the sample size). This is approximate as it is treating the sample as though it is sampled with replacement.

So the z-score of a mean of 41.7 is (41.7-41.26)/0.224, and the required odds of seeing a z-score less than this are:

P(z<(41.7-41.26)/0.224)/(1-P(z<(41.7-41.26)/0.224))

where P(.) is the cumulative standard normal probability function.

CB