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Diano4ka-milaya [45]
3 years ago
10

Answers matching "The combined average weight of an okapi and a llama is 450 kilograms. The average weight of 3 llamas is 190 ki

lograms more than the average weight of one okapi. On average, how much does an okapi weigh, and how much does a llama weigh?"
Mathematics
1 answer:
kirill [66]3 years ago
5 0

Let us say that:

x = weight of llamas

y = weight of okapi

 

From the problem, we can create the equations:

x + y = 450           --> 1

3 x = y + 190        --> 2

 

Rewriting equation 1:

x = 450 – y

 

From equation 2:

3 (450 – y) = y + 190

1350 – 3 y = y + 190

4 y = 1160

y = 290

 

From equation 1:

x = 450 – 290

x = 160

 

Therefore a llama weighs 160 kilograms while okapi weigh 290 kilograms on average.

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A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. The number of subjects needed to estima
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Answer:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

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.

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

ME represent the margin of error

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

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

2) Solution to the problem

Since the new 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 table 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

Assuming that the deviation is known we can express the margin of error is given by this formula:

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

And on this case we have that ME =\pm 3 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

Replacing into formula (b) we got:

n=(\frac{1.64(14.7)}{3})^2 =64.57 \approx 65

So the answer for this case would be n=65 rounded up to the nearest integer

And the sample size would decrease by 160-65=95 subjects

8 0
3 years ago
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