Question

A fire 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

Answer 1

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

Answer 2

Answer:

Option C) 0.25

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 22 minutes

Standard Deviation, σ = 11.9 minutes

We are given that the distribution of  response time is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(response times is over 30 minutes)

P(x > 30)

$P( x > 30) = P( z > \displaystyle\frac{30 - 22}{11.9}) = P(z > 0.6722)$

$= 1 - P(z \leq 0.6722)$

Calculation the value from standard normal z table, we have,  

$P(x > 30) = 1 - 0.749= 0.251 = 25.1\%$

Thus, the correct answer is

Option C) 0.25