Answer:0.0499
Step-by-step explanation:
Let x be a random variable that represents the salary for a data scientist.
Given: Average salary for a data scientist [tex]\mu=\$83105[/tex]
standard deviation [tex]\sigma=\$10503[/tex]
Sample size : n= 31
The probability that the mean salary of 31 randomly selected data scientists is less than $80,000.
[tex]P(\overline{X}<80000)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{80000-83105}{\dfrac{10503}{\sqrt{31}}})\\\\=P(z<-1.646)\ \ \ [z=(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(z<1.646)\ \ \ [P(Z<-z)=1-P(Z<z)]\\\\=1-0.9501= 0.0499[/tex]
Hence, the required probability = 0.0499.
