Answer:d is the correct option.
Step-by-step explanation:
Since its response time to fires is approximately normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = points scored by students
u = mean response time
s = standard deviation
From the information given,
u = 22 minutes
s = 11.9 minutes
We want to find the proportion of their response times that is over 30 minutes. It is expressed as
P(x > 30) = 1 - P(x ≤ 30)
For x = 30
z = (30 - 22)/11.9 = 0.67
Looking at the normal distribution table, the probability corresponding to the z score is 0.7486
P(x > 30) = 1 - 0.7486 = 0.25