Answer: 0.129
Step-by-step explanation:
Let [tex]\overline{X}[/tex] denotes a random variable that represents the mean weight of babies born.
Population mean : [tex]\mu= \text{3316 grams,}[/tex]
Standard deviation: [tex]\text{324 grams}[/tex]
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 :
[tex]P(|\mu-\overline{X}|>54)=1-P(\dfrac{-54}{\dfrac{324}{\sqrt{83}}}<\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{-54}{\dfrac{324}{\sqrt{83}}})\\\\=1-[P(-1.518<Z<1.518)\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{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[/tex]
hence, the required probability = 0.129
