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A bank manager wants to determine the percent of time the bank's tellers are working and idle. He decides to use work sampling, and his initial estimate is that the tellers are idle 20% of the time. Approximately how many observations should be taken to be 95% confident that the results will not be more than 5% away from the true result?

Answer :

cjmejiab

Answer:

246

Step-by-step explanation:

In the statment is defined our Values.

First que have a 95% Confidence interval

That is exactly in a Standar normal distribution table equal to Z Score = 1.96

The proportion [tex]\hat{p}[/tex] = 0.20

And we can and Margin Error (ME) equal to 0.05

Using the formula for the Sample Size (N) through Proportion we have,

[tex]N= \frac{Z_{score}^2*\hat{p}*(1-\hat{p})} {ME^2}[/tex]

[tex]N= \frac{1.96^2* 0.20 * (1-0.20)}{0.05^2}[/tex]

[tex]N= 245.86[/tex] or 246

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