The number of nuts in a package of "Premium Cashews" is normally distributed with a mean of 433 and a standard deviation of 6 nuts. Packages with fewer than 420 nuts or more than 445 nuts will be rejected by quality control.
a) what is the probability that a package selected at random has at most 430 nuts?
Attempt:
(430.5 - 433) / 6 = -0.42
using area under normal distribution curve. -0.42 = 0.3372
therefore there is a 33.72% chance? Am i right?
Thanks in advance!
I would not agree that this is correct. In particular, if $X$ ~ $N(\mu, \sigma^2)$ with pdf $f(x)$, and cdf $F(x)$, then the effect of the quality control mechanism is to doubly truncate the distribution, so that post-quality control, the distribution of nuts is say $g(x)$:
$$g(x) = \frac{f(x)}{F(b)-F(a)}$$
where $b=445$ is the upper bound, and $a=420$ is the lower bound, with domain of support on $(a,b)$, as illustrated here:
If $G(x)$ denotes the corresponding cdf, then the probability that a package selected at random has at most 430 nuts is $G(430) = P_{g}(X\le 430)$, which according to mathStatica/Mathematica is exactly:
which is approximately 0.304959 ... not 0.3372.
As an aside, the number of nuts in a package is discrete (unless perhaps you count the crumbles), whereas the Normal is continuous. It might very well be perfectly fine as an approximation, but that is not the structure that the question asserts.