Question

a study of nickels showed that the standard deviation of the weight of nickels is 300 milligrams. A coin counter manufacture wishes to find the 90% confidence interval for the average weight of a nickel. What is the minimum number of nickels he needs to weigh to obtain an average accurate to within 20 milligrams?

Answer 1

Using the z-distribution, a sample of 609 nickels has to be weighed.

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

For this problem, the parameters are:

$M = 20, \sigma = 300, z = 1.645$

We solve for n to find the sample size, then:

$M = z\frac{\sigma}{\sqrt{n}}$

$20 = 1.645\frac{300}{\sqrt{n}}$

$20\sqrt{n} = 1.645 \times 300$

$\sqrt{n} = 1.645 \times 15$

$(\sqrt{n})^2 = (1.645 \times 15)^2$

n = 608.9

Rounding up, a sample of 609 nickels has to be weighed.

More can be learned about the z-distribution at brainly.com/question/25890103

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