Answer: 0.0228
Step-by-step explanation:
Given : The mean and the standard deviation of finish times (in minutes) for this event are respectively as :-
[tex]\mu=30\\\\\sigma=5.5[/tex]
If the distribution of finish times is approximately bell-shaped and symmetric, then it must be normally distributed.
Let X be the random variable that represents the finish times for this event.
z score : [tex]z=\dfrac{x-\mu}{\sigma}[/tex]
[tex]z=\dfrac{19-30}{5.5}=-2[/tex]
Now, the probability of runners who finish in under 19 minutes by using standard normal distribution table :-
[tex]P(X<19)=P(z<-2)=0.0227501\approx0.0228[/tex]
Hence, the approximate proportion of runners who finish in under 19 minutes = 0.0228