Answer: 0.7619
Step-by-step explanation:
Given : Mean : [tex]\mu=45.0 [/tex]
Standard deviation : [tex]\sigma =8.0[/tex]
Sample size : [tex]n=9[/tex]
We assume that the variable is normally distributed.
The value of z-score is given by :-
[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
a) For x= 46.9 years
[tex]z=\dfrac{46.9-45.0}{\dfrac{8}{\sqrt{9}}}=0.7125[/tex]
The p-value : [tex]P(z<0.7125)=0.7619224\approx0.7619[/tex]
Hence, the  probability that the average age of those doctors is less than 46.9 years =0.7619
