Of 375 randomly selected students, 30 said that they planned to work in a rural community. Find 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

The 95% confidence interval for the true proportion of all medical students who plan to work in a rural community is (0.0525, 0.1075).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

95% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the true proportion of all medical students who plan to work in a rural community is (0.0525, 0.1075).

total amount to be paid = $19.5

step-by-step explanation:

given ,

original price = $25.99

discount = 25%

discount amount = 25% of original price

discount amount = 25/100   * 25.99

discount amount = $6.49

hence , the total amount to be paid = original price - discount amount

                                                            = $25.99 - $6.49

                                                            =$19.5

total amount to be paid = $19.5

6mm^3 or .6cm^3

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6     6   6

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