Normal Approximation to the Binomial Dist

[Bishop]

More than 200 billion grocery coupons are distributed each year for discounts exceeding $84 billion. However, according to a report in USA Today, only 3.2% of the coupons are redeemed. If a company distributes 5000 coupons, what is the probability that:


(a) at least 100 coupons are redeemed?

X>=100
N=5000 P= .0320
U=N*P= 5000 * .0320 = 160
Std Dev= Sqrt5000(.032)(.968) = 13.07
Z= 100 – 160/13.07 = -4.59
P(x >= 100) = P(z >= -4.59)

Now the problem I have is the z answer is well outside of the z chart to get the answer. Can someone please tell me what I did wrong?

 

[fencing]

I don't think you did anything wrong....P(z >= -4.59) is very close to 1 (you might choose the closest value to -4.59 you have in the chart...), so 1 is your answer. It makes intuitive sense: on average we would redeem 160 coupons with quite low standard deviation...95% of the data under normal assumptions would lie in +- 2*stdev window of the mean which is (190,130) in your case. Clearly 100 is way below...

 

[Bishop]

Thank you, but looking at the z chart it says anything over -3.9 should use .0000. Still confusing because this doesn't make sense.

The same goes for this next problem

fewer than 100 coupons are not redeemed? <

X < 100
N=5000 P= .968
U=N*P= 5000 * .968 = 4840
S= Sqrt5000 (.968) (.032) = 2.23
Z= x-u/o = 100 – 4840/2.23 = -2125.56 <--- this number is ridiculous....something must be wrong
P(x < 100) = P(z < )
=

Please anyone with any comments please feel free to help. Thanks

 

[ssd]

P(z>-4.59)= 0.99999778377. This is correct upto 10 digits and the 11th digit is rounded from the 12th digit. For such accurate values either you need a table used for post graduate classes or write a precise computer program.