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Math 80 – Quiz #10 (Chapter 18)
1. It is believed that nearsightedness affects about 14% of children. A kindergarten has registered
176 incoming children. Assuming the checks are satisfied,
a) Identify what p is in this problem. Write it in words and say what number it is.
p = 14%
p is the percent of all children who are affected by nearsightedness.
b) Identify what n is in this problem. Just say what numerical value it has.
n = 176
c) Calculate the mean and the standard deviation of the sample proportion.
The mean is p = 14% or
 ( pˆ )  p  0.14  14%
The standard deviation is
 ( pˆ ) 
(.14)(.86)
0.1204

 0.00068409  0.02615513 ... or 2.62%
176
176
d) About 95% of samples should obtain sample proportions between what two values?
About 95% of samples should be within about 2 (or 1.96) standard deviations of the mean.
So the two values are either:
0.14  2(0.262 )
14%  5.24%
8.76%  p  19.24%
OR
0.14  1.96(0.262)
14%  5.1352 %
8.8648 %  p  19.1352 %
Or you could just draw the model with 14% in the middle and go out 2 standard deviations
in both directions to find the two values.
e) What is the probability that over 30 of the incoming students are nearsighted?
First, notice that 30 out of 176 is approximately 17% (30/176). So, we are looking for the
probability that this group of 176 students will have more than 17% students who are
nearsighted. This means that pˆ  17% and we are looking for the probability that
pˆ  17% .
So, z 
0.17  0.14
 1.145...  1.15 . Using the chart, we can find that the percent of
0.262
data larger than 1.15 standard deviations above the mean in general is 12.51%, which
means that there is a 12.51% chance that our sample will have more than 17% of the
children affected by nearsightedness.