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Jobisdone [24]
3 years ago
8

at a high school, the probability that a student is a senior is 0.25. the probability that a student plays a sport is 0.20. the

probability that a student is a senior and plays a sport is 0.08. what is the probability that a randomly selected student plays a sport, given that the student is a senior?
Mathematics
2 answers:
makkiz [27]3 years ago
8 0
What are the answer to chose from
vladimir2022 [97]3 years ago
8 0

Answer:

0.32

Step-by-step explanation:

We have been given that at a high school, the probability that a student is a senior is 0.25. The probability that a student plays a sport is 0.20. The probability that a student is a senior and plays a sport is 0.08.

We will use conditional probability formula to solve our given problem. P(B|A)=\frac{P(\text{A and B)}}{P(A)}, where,

P(B|A) = The probability of event B given event A.

P(\text{A and B)} = The probability of event A and event B.

P(A) =Probability of event A.

Let A be that the student is senior and B be the student plays a sport.  

P(A and B) = Probability that student is a senior and plays a sport.

P(B|A)=\frac{\text{Probability that a student is a senior and plays a sport}}{\text{Probability that a student is senior}}

Upon substituting our given values we will get,

P(B|A)=\frac{0.08}{0.25}

P(B|A)=0.32

Therefore, the probability that a randomly selected student plays a sport, given that the student is a senior will be 0.32.


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A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the
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Answer:

a) \hat p=\frac{471}{1024}=0.460

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And the margin of error is given by:

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b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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Data given and notation  

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X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

\hat p=\frac{471}{1024}=0.460 estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering    

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

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Part b

If we replace the values obtained we got:

0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429

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