Suppose it is know that sigma is equal to 30. What is the minimum number of people the pollster should survey in order to have a margin of error of 10 miniutes at 95% confidence? show work.

To determine the minimum number of people the pollster should survey in order to have a margin of error of 10 minutes at 95% confidence, we can use the following formula:

n = (Z^2 * p * (1 - p)) / (E^2)

where:

n = the minimum number of people the pollster should survey

Z = the z-score corresponding to the desired level of confidence (for a 95% confidence level, the z-score is approximately 1.96)

p = the estimated proportion of the population that holds a particular opinion or characteristic (in this case, we will assume that p = 0.5, since we are given that sigma is equal to 30 and we want to find the minimum number of people needed to achieve a margin of error of 10 minutes)

E = the desired margin of error (in this case, E = 10 minutes)

Substituting these values into the formula, we get:

n = (1.96^2 * 0.5 * (1 - 0.5)) / (10^2)

Simplifying this expression, we get:

n = (3.8416 * 0.25) / 100

= 0.09604 / 100

= 0.0009604

Since n must be a whole number, we need to round up to the nearest whole number. In this case, the minimum number of people the pollster should survey is 1.

Therefore, the minimum number of people the pollster should survey in order to have a margin of error of 10 minutes at 95% confidence is 1.

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