In a city of over 1,000,000 residents, 14% of the residents are senior citizens. In a simple random sample of 1200 residents, there is about a 95% chance that the percent of senior citizens is in the interval [pick the best option; even if you can provide a sharper answer than you see in the choices, please just pick the best among the options a-d]
9% to 19% 10% to 18% 11% to 17% 12% to 16% 13% to 15%

Question

In a city of over 1,000,000 residents, 14% of the residents are senior citizens. In a simple random sample of 1200 residents, there is about a 95% chance that the percent of senior citizens is in the interval [pick the best option; even if you can provide a sharper answer than you see in the choices, please just pick the best among the options a-d]
9% to 19% 10% to 18% 11% to 17% 12% to 16% 13% to 15%

anonymous · Answer

can any one pls reply

anonymous · Answer

A simple random sample of 50 students is taken from a class of 300 students. In the class,

 * the average midterm score is 67 and the SD is 12

* there are 72 women

Let W be the number of women in the sample, and let S be the average midterm score of the sampled students.
find E(w)
SE(W),E(S),SE(S)

anonymous · Answer

PLS any one can reply fr the above two questions plss

anonymous · Answer

I need insert on them

anonymous · Answer

well dinkar some time you need to give answer

kropot72 · Answer

The first question can be solved by using the Normal approximation to the binomial distribution.
n = 1200
p = 0.14
$$mean=np=1200\times 0.14=168$$
$$standard\ deviation=\sqrt{np(1-p)}=\sqrt{144.48}=12.02$$
The empirical rule for a Normal distribution states that approximately 95% of the data points lie within the range plus and minus 2 standard deviations of the mean.
Therefore the lower limit = 168 - (2 * 12) = 144
and the upper limit = 168 + )2 * 12) = 192
144 is (144/1200) * 100 = 12% of the sample.
192 is (192/1200) * 100 = 16% of the sample.