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It is generally believed that nearsightedness affects about 15% of children.  A school district gives vision tests...

It is generally believed that nearsightedness affects about 15% of children.  A school district gives vision tests to 111 incoming kindergarten children. Use the empirical rule (68%-95%-99.7% Rule) to determine what proportion of nearsighted children we might expect to see in samples of 111 children (I'm not looking for the number of children).  Assume conditions are met!

Based on your results, would you be surprised to find a sample where 20% of children were nearsighted? Find the z-score and resulting probability to make this determination, rather than just analyzing the number of standard deviations.

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Answer #1
for normal distribution z score =(p̂-p)/σp
here population proportion=     p= 0.150
sample size       =n= 111
std error of proportion=σp=√(p*(1-p)/n)= 0.0339

hence probability of having 20% or more children nearsighted :

probability = P(X>0.2) = P(Z>1.48)= 1-P(Z<1.48)= 1-0.9306= 0.0694

as this probability is not less then 0.05 level of unusual events ; therefore this is not an unusual event

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