The Federal Government is stepping up efforts to reduce average response times of fire departments to fire calls. Nationwide, the distribution of average response times to fire calls follows a normal distribution with a mean of 12.8 minutes and a standard deviation of 3.7 minutes. The fastest 20% of fire departments will be singled out for a special safety award. How fast must a fire department be in order to qualify for the special safety award

Answer: in order to qualify, the time must be at least 15.93 minutes

Step-by-step explanation:

Let x be a random variable representing the average response times of fire departments to fire calls. Since it follows a normal distribution, we will determine the z score by applying the formula,

z = (x - µ) / σ

Where

x = sample mean

µ = population mean

σ = population standard deviation

From the information given,

µ = 12.8 minutes

σ = 3.7 minutes

in order to qualify for the special safety award, the probability value would be to the right of the z score corresponding to 80%

The z score corresponding to 80% on the normal distribution table is 0.845

Therefore,

0.845 = (x - 12.8) / 3.7

0.845 * 3.7 = x - 12.8

3.13 = x - 12.8

x = 12.8 + 3.13 = 15.93