Answer: 0.16
Step-by-step explanation:
Given that the run times provided are normally distributed ;
Mean(x) of distribution = 3 hours 50 minutes
Standard deviation(s) = 30 minutes
The probability that a randomly selected runner has a time less than or equal to 3 hours 20 minutes
3 hours 20 minutes = (3 hrs 50 mins - 30 mins):
This is equivalent to :
[mean(x) - 1 standard deviation]
z 1 standard deviation within the mean = 0.84
z, 1 standard deviation outside the mean equals:
P(1 - z value , 1standard deviation within the mean)
1 - 0.8413 = 0.1587
= 0.16
