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3 years ago

What is 5 over20 by simplifying fraction

Mathematics

2 answers:

inna [77]3 years ago

6 0

5 over 20 simplifies to 1 over 4

babymother [125]3 years ago

4 0

To simplify a fraction, divide numerator and denominator by
their greatest common factor.

The greatest common factor of 5 and 20 is 5 .

5/20 = 1 / 4

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An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

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a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that $n = 1600, \pi = \frac{8}{1600} = 0.005$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095$

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A sample of n is required, and n is found for M = 0.009. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}$

$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2$

$n = 407.3$

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use $\pi = 0.5$