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oee [108]
3 years ago
13

www.g A survey of athletes at a high school is conducted, and the following facts are discovered: 19% of the athletes are footba

ll players, 79% are basketball players, and 14% of the athletes play both football and basketball. An athlete is chosen at random from the high school: what is the probability that they are either a football player or a basketball player
Mathematics
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

84%

Step-by-step explanation:

The probability that the selected player is a football player, P(F)=19%

The probability that the selected player is a basketball player, P(B)=79%

The probability that the selected player play both football and basketball,

P(B \cap F)=14\%

We want to determine the probability that a randomly chosen athlete is either a football player or a basketball player, P(B \cup F)

In probability theory

P(B \cup F)=P(B)+P(F)-P(B \cap F)\\=79\%+19\%-14\%\\=84\%

The probability that a randomly chosen athlete is either a football player or a basketball player is 84%.

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A manufacturer reported a sample mean = 22.0 g and a sample standard deviation = 2.5 g based on a sample of 20 of their products
tester [92]

Answer:

n=(\frac{1.640(2.5)}{2})^2 =4.2 \approx 5

So the answer for this case would be n=5 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

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.

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".

\bar X=22 represent the sample mean

\mu population mean (variable of interest)

s=2.5 represent the sample standard deviation

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Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (2)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (3)

We can use as estimator for the population deviation the sample deviation \hat \sigma = s

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got z_{\alpha/2}=1.640, replacing into formula (3) we got:

n=(\frac{1.640(2.5)}{2})^2 =4.2 \approx 5

So the answer for this case would be n=5 rounded up to the nearest integer

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Answer:

Please check explanations

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