The normal distribution?

Question

97 out of 100 people cannot detect the difference in taste between new(healthier) and old cooking oils. In a random sample of 1000 individuals, what is the probability of

a) at least 40 can taste the difference between the oils?

b) at most 5% can taste the difference between the oils?

Please help me!

Answer 1

Here you want to use the normal approximation to the binomial F_Y:

F_Y(t) ~ N_z((t - np)/sqrt(npq))

where N_z is the standard normal; n is the number in the sample, which here is 1000; p is the probability of success, which here is 0.03; and q=1-p.

So np = 30 and npq = 29.1

Then part a is

1 - N_z((40 - 30)/sqrt(29.1))
= 1 - N_z(1.854) = 1 - 0.9681 = 0.0319

or if you use "continuity correction" it's

1 - N_z((39.5 - 30)/sqrt(29.1))
= 1 - N_z(1.761) = 1 - 0.9609 = 0.0391

(It's 1 minus the normal because the question asks for the probability of 40 or more. That's also why I subtracted 0.5 from 40 in the "continuity correction" calculation.)

Part b is very similar, and I leave it to you. Since it's "at most",
it's just the normal, not 1 minus the normal.