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k0ka [10]
3 years ago
6

Mr. Foley wants to randomly select 3 students for a committee. He used 3 coins to conduct a simulation to predict the probabilit

y that the committee will have at least 2 girls. The results of 16 trials of the simulation are shown below. Let H represent a girl and T represent a boy.
TTH HTH THH HHH
HTT THT THT HHT
HHT THH TTT HTT
HHH HTH TTH HHT
Based on the results of the simulation, what is the probability that the committee Mr. Foley selects will have at LEAST 2 girls?
Mathematics
1 answer:
jek_recluse [69]3 years ago
4 0
The probability is 9/16.

This is because in the simulation, 9 of the results had 2 or 3 girls, and there was a total of 16 choices.
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The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,
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Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) 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.

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\bar X=400 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=80 represent the population standard deviation

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2) Part a

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that z_{\alpha/2}=2.58

Now we have everything in order to replace into formula (1):

400-2.58\frac{80}{\sqrt{15}}=346.708    

400+2.58\frac{80}{\sqrt{15}}=453.292

So on this case the 99% confidence interval would be given by (346.708;453.292)    

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

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

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

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that t_{\alpha/2}=2.98

Now we have everything in order to replace into formula (1):

400-2.98\frac{80}{\sqrt{15}}=338.445    

400+2.98\frac{80}{\sqrt{15}}=461.555

So on this case the 99% confidence interval would be given by (338.445;461.555)    

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