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gizmo_the_mogwai [7]
3 years ago
12

Please help me Question: What are the solutions to this graph?

Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
8 0
There’s only one solution. Solution is the same as x or x intercepts so it would be one at (2,0)
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According to a research paper, the carbon monoxide exposure of someone riding a motorbike for 5 km on a highway in a particular
KatRina [158]

Answer:

a) P(X>20)=P(Z>0.271)=1-P(Z

b) P(X>25)=PP(Z>1.119)=1-P(Z

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

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

6 0
3 years ago
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