You want to know about the dining habits of college students in the US. Since you do not have resources to survey all college students in the nation, you take a random sample of 1501. You learn that, on average, they eat out 4 times per week. Also, these sample data are generally distributed about the average of 4 times, by 1.5 times per week. 1. What is the average times per week that college students across the nation eat out

the average times per week that college students across the nation eat out lies within these 95% confidence interval

Step-by-step explanation:

From the question we are told that

    The sample size is n =  1501

    The sample mean  is  

      The standard deviation is  

From the question we are told the confidence level is  95% , hence the level of significance is    

=>  

Generally from the normal distribution table the critical value  of   is  

Generally the margin of error is mathematically represented as  

=>  

=>  

Generally the estimate for average times per week that college students across the nation eat out at a 95% confidence is  mathematically represented as

=>  

Hence the true average times per week that college students across the nation eat out lies within this interval

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step-by-step explanation:

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*