Answer: Choice C) 0.6964; 0.6972
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Work Shown:
n = 15
p = 0.5
q = 1-p = 1-0.5 = 0.5
P(k) = (n C k)*(p)^k*(q)^(n-k)
P(k) = (15 C k)*(0.5)^k*(0.5)^(15-k)
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If you plug in k = 7, then we get,
P(k) = (15 C k)*(0.5)^k*(0.5)^(15-k)
P(7) = (15 C 7)*(0.5)^7*(0.5)^(15-7)
P(7) = 6435*(0.5)^7*(0.5)^8
P(7) = 0.1964
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Repeat for k = 8 all the way through k = 15. You could do this by hand, but I recommend using a spreadsheet to make things go much quicker.
Once you determine those values, add them up and you should get 0.6964 which is the binomial probability we want.
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As for the normal approximation, you'll need to compute the mu and sigma to get
- mu = n*p = 15*0.5 = 7.5
- sigma = sqrt(n*p*q) = sqrt(15*0.5*0.5) = 1.93649 which is approximate.
The normal distribution will have those parameters.
Since we're using a continuity correction, we need to bump the x = 6 up to x = 6.5, since we want to be larger than 6
Let's find the z score
z = (x - mu)/sigma
z = (6.5 - 7.5)/(1.93649)
z = -0.516398
Now use a Z table or a calculator to determine that P(Z > -0.516398) = 0.6972 approximately. If you're using a TI calculator, then you'll use the normalCDF function. If you're using excel, then you would use the NormDist function (make sure to turn the cumulative flag to "true"). Alternatively, you can search out free z calculators to get the job done.
