Question

a fire department in a rural county reports that its response time to fires is approximately normally distributed with a mean of 22 minutes and a standard deviation of 6.9 minutes. assume that this claim is true.one percent of response times take at least how many minutes?

Answer 1

If this claim is true then one percent of response times take at least $38.077$ minutes.

In the given question,

A fire department in a rural county reports that its response time to fires is approximately normally distributed with a mean of 22 minutes and a standard deviation of 6.9 minutes.

We have to assuming that the claim is true and finding that one percent of response times take at least how many minutes.

From the question,

Mean $\mu=22$ minutes

Standard deviation $\sigma=6.9$ minutes.

Since the given percentage is $1\%$ and percentage means division of $100$ so we can write $1\%=1/100=0.01$

So the one percent of response times take at least how many minutes is

$P(x\geq a)=0.01$

$P(x\geq \frac{a-\mu}{\sigma})=0.01$

As we know the all values. SO putting the values in the formula

$P(x\geq \frac{a-22}{6.9})=0.01\dots\dots$Equation 1

From the z table

$P(z\geq 2.33)=0.01\dots\dots$Equation 2

Now from Equation 1 and Equation 2

$\frac{a-22}{6.9}=2.33$

Multiply by $6.9$ on both side

$\frac{a-22}{6.9}\times6.9=2.33\times6.9$

Simplifying

$a-22=2.33\times6.9$

$a-22=16.077$

Add $22$ on both side

$a-22+22=16.077+22$

$a=38.077$

Hence, if this claim is true then one percent of response times take at least $38.077$ minutes.

To learn more about standard deviation link is here

https://brainly.com/question/14747159

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