The mean percentange of a population of people eating out at least once a week is 57℅

The correct answer is B. between 56.45% and 57.55%

Complete statement and question:

The mean percentage of a population of people eating out at least once a week is 57% with a standard deviation of 3.50%. Assume that a sample size of 40 people was surveyed from the population a significant number of times. In which interval will 68% of the sample means occur?

between 55.89% and 58.11%

between 56.45% and 57.55%

between 56.54% and 57.46%

between 56.07% and 57.93%

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Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Mean percentage of a population of people eating out at least once a week  = 57%

Standard deviation = 3.5%

Sample size = 40

Confidence level = 68%

2. In which interval will 68% of the sample means occur?

For answering this question, we should find out the standard deviation of the sample, using this formula:

Standard deviation of the sample = Standard deviation of the population/√Sample size

Standard deviation of the sample = 3.5/√40

Standard deviation of the sample = 3.5/6.32

Standard deviation of the sample = 0.55

Let's recall that a confidence level of 68% means that 68% of the sample data would have a value between the mean - 1 time the standard deviation of the sample and the mean  + 1 time the standard deviation of the sample. Thus:

57 - 1 * 0.55 = 57 - 0.55 = 56.45

57 + 1  * 0.55 = 57 + 0.55 = 57.55

The correct answer is B. between 56.45% and 57.55%

cosine

step-by-step explanation: