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4 years ago

In a particular class of 27 students,17 are men.what fraction of the student in the class are men

2 answers:

8 0

$p \% \ of \ 27=17 \\\\ \frac{p}{100}*27=17 \\\\ \frac{p}{100}=\frac{17}{27} \\\\ p=\frac{100*17}{27} \\\\ \boxed{p=\frac{1700}{27}}$

igomit [66]4 years ago

6 0

Total students = 27
total men = 17
Now,
Fraction of man in class = total men / total students
= 17 / 27
Since, the fraction can't be further simplified, the answer is $\frac{17}{27}$

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Answer:

$652.6-2.01\frac{311.7}{\sqrt{50}}=563.997$

$652.6+2.01\frac{311.7}{\sqrt{50}}=741.203$

So on this case the 95% confidence interval would be given by (563.997;741.203)

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".

$\bar X=652.6$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=311.7 represent the sample standard deviation

n=50 represent the sample size

Soltuion 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)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=50-1=49$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,49)".And we see that $t_{\alpha/2}=2.01$