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

(0.05256, 0.10744) .

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

  (1)

, where = Sample proportion

n= Sample size

z* = Critical z-value.

Let p be the population proportion of all medical students who plan to work in a rural community.

As per given , we have

n= 375

Critical z-value for 95% confidence interval = 1.96

Put all values in (1) , we get

Hence, the 95% confidence interval for the true population proportion of all medical students who plan to work in a rural community is (0.05256, 0.10744) .

answer: 230 miles ( vote ! )

step-by-step explanation:

180.39 minus 17.99 equals 162.4

then divide 162.4 by .80 to get 230

me! what did you put for this question?

step-by-step explanation: