Question

According to a previous year's report, 18% of children have myopia (the condition of nearsightedness). A random sample of 195 children recently had their vision checked Based on the sample of 195 children, show that the SuccessFailure condition is satisfied so that the sampling distribution of is approximately normal. Don't round the results. b) Based on the sample of 195 children, find the mean and standard deviation of the sampling distribution of p. Round the standard deviation to 4 decimal places. c) From the sample of 195 children, what is the probability that the proportion of them having myopia is above 227 Hint Find P DO NOT ROUND the final answer. >0.22). Round your scores to 2 decimal places

Answer 1

Å shivena p r oportion of childien huwing --- myopia X: # of children having myopia we know that sompling distribution of is approximately normal if - np > 10 & ng > 10 n = 155 -- p: 0:18.- ...np 35:1 -0 _ ng - 159. 9 0 as both conditions of 6 are saisfied => ê ñ N/P Å ã N/0.18 Pli-p? where p = 0.18 0.1176) n=195 195 ELÊ = p = 0.18 VCÊ = P(12) 01476 195

000275: (قال : (ماله .ق) ليفه تنها تفاحه (.ح ) م .: 0 . 18 - 0 . 22 د ط اعت 0:00 21S = 5387 ہا : 72] (38 5 با- ه - ل 61ق.م -ا - ۵۰ 01353 - 7353 همت (22 00 حمام