We know that, in this case, the sample mean will be equal to the population mean, so
[tex]\mu_{\bar{x}}=250gr[/tex]The standard deviation of the sample follows the following formula
[tex]\sigma_{\bar{x}}=\frac{\sigma}{\sqrt[]{n}}=\frac{42gr}{\sqrt[]{20}}\approx9.4[/tex]The sample standard deviation is 9.4, approximately.
At last, to find the probability of the sample, first, we have to find the z-score using the following equation
[tex]z=\frac{x-\mu_{\bar{x}}}{\sigma_{\bar{x}}}=\frac{38.5-250}{9.4}=\frac{-211.5}{9.4}=-22.5[/tex]The probability value assigned to z = -22.5 is near to 0, which means that it is almost sure that there's no sample mean less than 38.5 grams.