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Andreyy89
2 years ago
7

A group of people were asked if they had run a red light in the last year where 140 responded "yes," and 208 responded "no." If

a person is randomly chosen, find the probability that he/she has run a red light in the last year. Give your answer as a fraction or decimal.
Mathematics
1 answer:
gayaneshka [121]2 years ago
5 0

Answer:

35/87

Step-by-step explanation:

To find the probability, we must find (total number of favorable outcomes)/ (total number of outcomes). We thus need to find the total number of outcomes. As people have only answered "yes" or "no", the total number of outcomes is 208+140 = 348.

The total number of favorable outcomes, or what we're looking for, is 140 (140 people responded that they had run a red light). Thus, our fraction is 140/348, or 35/87

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Answer:

a) P(R

And we can find this probability using the normal standard table or excel and we got:

P(z

b) P(R>110)=P(\frac{R-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{3.606})=P(Z>2.774)

And we can find this probability using the complement rule and the normal standard table or excel and we got:

P(z>2.774)=1-P(Z

Step-by-step explanation:

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".  

Part a

Let X the random variable that represent the time for the step 1 and Y the time for the step 2, we define the random variable R= X+Y for the total time and the distribution for R assuming independence between X and Y is:

R \sim N(40+60 = 100,\sqrt{2^2 +3^2}= 3.606 s)  

Where \mu=65.5 and \sigma=2.6

We are interested on this probability

P(R

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

z=\frac{R-\mu}{\sigma}

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

P(R

And we can find this probability using the normal standard table or excel and we got:

P(z

Part b

P(R>110)=P(\frac{R-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{3.606})=P(Z>2.774)

And we can find this probability using the complement rule and the normal standard table or excel and we got:

P(z>2.774)=1-P(Z

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