Can't figure out what I'm supposed to do here.

[Zilliano]

The math for stats isn't all that hard, but my teacher has a way of confusing us by contradicting himself to the point where we no longer know what we're supposed to do on a problem. Argh.

Anyway, here's the problem:

A seed wholesaler tests his tomato seed and calculates their germination rate to be 20 per package of 25 with a standard deviation of 2. He offers a money-back guarantee if less than 15 of his seed germinate. Consider a package of 25 tomato seeds, and answer the following questions.

A. If the wholesaler sells 2,872,115 packages of tomato seed, how many packages should he have to replace?

B. If he wishes to replace less than 2% of the seed packages under his money-back guarantee, at what germination rate should he advertise his money-back offer?

C. Due to competition, he feels he must offer a better warranty than his competitor who replaces 10% of his tomato seed. If he were to replace about 11% under warranty, what would his germination rate be per pack of 25 tomato seed?

I'm completely lost as to WHAT I'm supposed to apply to this problem.

Thanks in advance for any help.

 

[tkhunny]

A seed wholesaler tests his tomato seed and calculates their germination rate to be 20 per package of 25 with a standard deviation of 2. He offers a money-back guarantee if less than 15 of his seed germinate. Consider a package of 25 tomato seeds, and answer the following questions.

A. If the wholesaler sells 2,872,115 packages of tomato seed, how many packages should he have to replace?

Numbers (Normal Approximation)
\(\displaystyle \frac{15-20}{2}\;=\;-5/2\;= \;-2.5\)

How much of that "Normal" area is below 2.5 standard deviations below the mean? I get about 0.621%

OR

Porportions (Normal Approximation)
\(\displaystyle \frac{0.6-0.8}{0.08}\;=\;-0.2/0.08\;= \;-2.5\)

OR

Numbers (Binomial)

\(\displaystyle g('n'-will-germinate) = \frac{25!}{n!\cdot (25-n)!}\cdot \left(\frac{4}{5}\right)^{n}\cdot \left(\frac{1}{5}\right)^{25-n}\)

p(14 or fewer germinate) = g(0) + g(1) + g(2) + ... + g(14) = ?? I get about 0.556%. How does this compare to the previous "normal" results? Does the difference make the extra effort worth the time?

2872115 * (0.00621 - 0.00556) = 1867 packets. Is our friend willing to be that far off?