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ch4aika [34]
3 years ago
10

A prize was awarded to 24 women and 1130 men. a. What fraction of the prize winners were​ women? b. What fraction were​ men?

Mathematics
1 answer:
SpyIntel [72]3 years ago
7 0

A) The fraction of the prize winners were​ women is \frac{12}{577}

B) The fraction of the prize winners were​ men is \frac{565}{577}

<em><u>Solution:</u></em>

Given that, A prize was awarded to 24 women and 1130 men

Therefore, Women + men = 24 + 1130 = 1154

<em><u>A) What fraction of the prize winners were​ women?</u></em>

The fraction of the prize winners were​ women is found by dividing the total women by total men and women

\text{Fraction women } = \frac{\text{Number of women}}{\text{Total men and women}}

Total men and women = 1154

Number of women = 24

Thus, we get,

\text{Fraction women } = \frac{24}{1154}\\\\\text{Reduce to lowest terms }\\\\\text{Fraction women } = \frac{12}{577}

<em><u>b.) What fraction were​ men?</u></em>

The fraction of the prize winners were​ men is found by dividing the total men by total men and women

\text{Fraction women } = \frac{\text{Number of men}}{\text{Total men and women}}

Total men and women = 1154

Number of men = 1130

Thus, we get,

\text{Fraction men } = \frac{1130}{1154}\\\\\text{Reduce to lowest terms }\\\\\text{Fraction men } = \frac{565}{577}

Thus the fractions are found

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10.08-1.64\frac{0.1}{\sqrt{6}}=10.013    

10.08+1.64\frac{0.1}{\sqrt{6}}=10.147    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

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".

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

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

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=10.08

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that z_{\alpha/2}=1.64

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

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10.08+1.64\frac{0.1}{\sqrt{6}}=10.147    

So on this case the 90% confidence interval would be given by (10.013;10.147)    

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