Simplify (1/4 ad^3)^2

1 answer:

7 0

That "ad^3" bothers me; I'm not sure what you mean here.

I'm going to assume that you have (1/4*a*d^3)^2.

If that's the case, then you have (1/4)^2 * a^2 * d^6 = (1/16)*a^2*d^6.

If this is not what you were hoping for, double check to ensure that you have copied down this problem exactly as it was presented.

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Answer:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

Step-by-step explanation:

For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.

We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

$\hat p = \frac{24}{250}= 0.096$

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".

The confidence interval for a proportion is given by this formula

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

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.