Question

According to a research paper, the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in a particular city is approximately normally distributed with a mean of 18.4 ppm. Suppose that the standard deviation of carbon monoxide exposure is 5.9 ppm.

Answer 1

Answer:

a) $P(X>20)=P(Z>0.271)=1-P(Z<0.271)=1-0.6068=0.3932$

b) $P(X>25)=PP(Z>1.119)=1-P(Z<1.119)=1-0.8684=0.1316$

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Part a

Let X the random variable that represent the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in a particular city of a population, and for this case we know the distribution for X is given by:

$X \sim N(18.4,5.9)$  

Where $\mu=18.9$ and $\sigma=5.9$

(a) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monexide exposure of more than 20 ppm? (Round your answer to fou decimal places.)

We are interested on this probability

$P(X>20)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>20)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{20-18.4}{5.9})=P(Z>0.271)$

And we can find this probability on this way:

$P(Z>0.271)=1-P(Z<0.271)=1-0.6068=0.3932$

b) Approximately what proportion of those who ride a motorbike for 5 km on a highway in this city will experience a carbon monoxide exposure of more than 25 ppm? (Round your answer to four decimal places.)

We are interested on this probability

$P(X>25)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>25)=P(\frac{X-\mu}{\sigma}>\frac{20-\mu}{\sigma})=P(Z>\frac{25-18.4}{5.9})=P(Z>1.119)$

And we can find this probability on this way:

$P(Z>1.119)=1-P(Z<1.119)=1-0.8684=0.1316$