Answer: 0.129
Explanation:
Let
denotes a random variable that represents the mean weight of babies born.
Population mean :

Standard deviation:

Sample size = 83
Now, the probability that the mean weight of the sample babies would differ from the population mean by greater than 54 grams will be :
![P(|\mu-\overline{X}|>54)=1-P((-54)/((324)/(√(83)))<\frac{\overline{X}-\mu}{(\sigma)/(√(n))}<(-54)/((324)/(√(83))))\\\\=1-[P(-1.518<Z<1.518)\ \ \ [Z=\frac{\overline{X}-\mu}{(\sigma)/(√(n))}]\\\\=1-[P(Z<1.518)-P(z<-1.518)]\\\\=1-[P(Z<1.518)-(1-P(z<1.518))]\\\\=1-[2P(Z<1.518)-1]=2-2P(Z<1.518)\\\\=2-2(0.9355)\ [\text{By z-table}]\\\\=0.129](https://img.qammunity.org/2021/formulas/mathematics/college/edt0ly8rpri2n67fspan37cwkzxxv46nlf.png)
hence, the required probability = 0.129