Question

The owner of a shopping mall wants an estimate of the lengthof

Answer 1

Answer:

A sample size of at least 87 will be needed.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.01 = 0.99$, so $z = 2.325$

Now, find M as such

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

$M = 5, \sigma = 20$

So

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

$5 = 2.325*\frac{20}{\sqrt{n}}$

$5\sqrt{n} = 46.5$

$\sqrt{n} = 9.3$

$n = 86.49$

A sample size of at least 87 will be needed.