Afire department in a rural county reports that its response time to fires is approximatelynormally distributed with a mean of 22 minutes and a standard deviation of 11.9 minutes. approximately what proportion of their response times is over 30 minutes? a) 0.03b) 0.21c) 0.25d) 0.75e) 0.79

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

bn,mn,n.vn.

step-by-step explanation: