Ali bought 20 dozen oranges 10% of them were rotten. How many oranges were good ?

Mathematics

1 answer:

Mashutka [201]2 years ago

5 0

Answer:

Step-by-step explanation:

rotten oranges=10/100*20

200/100

2 dozen

good oranges=total dozen-rotten oranges

20-2

18 d0zen were good oranges.

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You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estim

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Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.025}{1.28})^2}=655.36$

And rounded up we have that n=656

Step-by-step explanation:

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 80% of confidence, our significance level would be given by $\alpha=1-0.80=0.20$ and $\alpha/2 =0.10$ . And the critical value would be given by:

$z_{\alpha/2}=\pm 1.28$

Solution to the problem

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

Since we don't have prior info for the proportion of interest we can use $\hat p=0.5$ as estimator. And on this case we have that $ME =\pm 0.025$ and we are interested in order to find the value of n, if we solve n from equation (a) we got: