Question 414381
It is estimated that an average of 53% of college students graduate in 5 years or less.  If 500 students on a large campus are selected at random, what is the probability that between 50% and 60% of them will graduate in 5 years or less?
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This is a binomial problem with

{{{n = 500}}}
{{{p = .53}}}

You are asked:

What is the probability that between 50% and 60% of them 
will graduate in 5 years or less?

Since 50% of 500 is 250 and 69% of 500 is 300, this is the
same as being asked

What is the probability that between 250 and 300 of the 500 
will graduate in 5 years or less?

You can do this on a TI-83 or TI-84 by

binomcdf(500,.53,300) - binomcdf(500,.53,250)

getting .902291559.

Or you can do it by approximating the binomial with the normal.

{{{n = 500}}}
{{{p = .53}}}
{{{mu = np = 500*.53 = 265}}}
{{{sigma = sqrt(np(1-p)) = sqrt(500*.53*(1-.53)) = 11.16019713}}}

normalcdf(249.5,300.5,265,11.16019713)

getting .916828909

Or if your teach won't let you use a calculator and insists on
less accurate tables, then find the z-scores of

249.5 and 300.5,

{{{z=(x-mu)/sigma}}}

{{{z=(249.5-265)/11.16019713}}}

{{{z=-1.39}}}

{{{z=(x-mu)/sigma}}}

{{{z=(300.5-265)/11.16019713}}}

{{{z=3.18}}}

and, depending on what kind of table you have, you will
either 

1. If your table has only positive z-scores,

look up 1.38, finding .4162, look up 3.18, finding .4993,
add these and get .9155

or 

2. If your table has both positive and negative z-scores,

look up -1.38, finding .0838, look up 3.18, finding .9993,
subtract these and get .9155.

Edwin</pre>