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) .

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

answer: 140

step-by-step explanation:

so just add all then it will be your answer but i am not pretty sur eif it is but hope this i am here to people