The times a fire department takes to arrive at the scene of an emergency are normally distributed with a mean of 6 minutes and a standard deviation of 1 minute. For about what percent of emergencies does the fire department arrive at the scene in between 4 minutes and 8 minutes ?

Step-by-step explanation:

Let x be the random variable representing the times a fire department takes to arrive at the scene of an emergency. Since the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 6 minutes

σ = 1 minute

the probability that fire department arrives at the scene in case of an emergency between 4 minutes and 8 minutes is expressed as

P(4 ≤ x ≤ 8)

For x = 4,

z = (4 - 6)/1 = - 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.023

For x = 8

z = (8 - 6)/1 = 2

Looking at the normal distribution table, the probability corresponding to the z score is 0.98

Therefore,

P(4 ≤ x ≤ 8) = 0.98 - 0.23 = 0.75

The percent of emergencies that the fire department arrive at the scene in between 4 minutes and 8 minutes is

0.75 × 100 = 75%

sup

step-by-step explanation:

answer: idk

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