Question 286508
An accounting professor claims that one quarter of undergraduate business students major in accounting. If we assume this is true, what is the probability that in a random sample of 790 undergraduate business students, 179 or more will major in accounting? 
<pre><font size = 4 color = "indigo"><b>
If you have a TI-83 or 84 calculator,

Press CLEAR
Type this:   1-
Press 2nd
Press VARS
Press ALPHA
Press APPS  
You will see this:    1-binomcdf(
Type this:            790,1/4,179)
You will see this:    1-binomcdf(790,1/4,179)
Press ENTER

You will see this:    .9315736737

If you don't have a calculator that will do this directly,
you must use a NORMAL APPROXIMATION OF THE BINOMIAL:

Calculate the mean: {{{mu=n*p=(790)(1/4)=197.5}}}

Calculate the standard deviation: 
{{{sigma=sqrt(n*p*(1-p))=sqrt(790*(1/4)*(1-1/4))=sqrt(790*.25*.75)}}}
{{{""=sqrt(148.125)=12.17066144}}}

We want to find P(x <u>></u> 179) Since the approximation 
is to include 179, and 179 is less than the mean, 
we want P(x > 179.5)

Calculate the z-score for 178.5

{{{z=(x-mu)/sigma=(179.5-197.5)/12.17066144=-1.478966454}}}

P(x > 179.5) translates to P(z > -1.478966454)

Round the absolute value of z to hundredths. That's 1.48

In the nomal table find 1.4 in the leftmost column.
Find the hundredths digit 8 (0.08) at the top, then read
0.4306 if you have the kind of table that reads from the
middle. Then add .5 to that and get 0.9306.  (If you have
the other kind of table you will read 0.9306 driectly by
looking up 1.48).

The approximation using the normal table is 0.9306.

The reason you don't get the same answer is because
the calculator is more accurate than to normal tables.
But you will noticethat both answers are .93 when
rounded to hundredths.

Edwin</pre>