prob. normal distribution

Question

Slices of pizza for a certain brand of pizza have a mass that is approximately normally distributed with a mean of 66.1 grams and a standard deviation of 1.97 grams.

c) What is the probability of finding a 16 random slices of pizza with a mean mass of less than 65.2 grams? 



d) What sample mean (for a sample of size 16) would represent the bottom 15% (the 15th percentile)?  grams

Answer 1

c)

A very straightforward question.

To answer it, we will have to assume that "probability of finding a16 random slices of pizza with ..." means "probability that an independent sample of 16 random slices of pizza is with ..."

Bear in mind that if two slices come from the same pizza, then their masses are not independent, because, for example, if you are a slice drawn from the same pizza as some extraordinarily massive slice, then you are extraordinarily likely to be a smaller than average piece. So if you want to do this experiment properly, use 16 pizzas from each of which you draw one slice at random. It's more expensive than, say, drawing 2 pizzas at random and using the eight slices of each. But the 16 pizza sample will be much closer to an independent sample.

Let Y be the random variable which is the average of 16 slices drawn randomly (all 8 slices equally likely to be drawn) from a random sample of 16 pizzas. Then Y is also normally distributed with mean 66.1 grams (same as mean of a random slice). And Y's standard deviation is 1/4 of the standard deviation (sd) of the of the sd of a single slice, since 4 = square root of the sample size of16. So the sd of Y is approximately 1.97 / 4 = 0.4925 grams .

Now 65.2 is less than Y's average by 66.1 - 65.2 = 0.9, and 0.9 is about 0.9 / 0.4925 = 1.827 of Y's sd's. So the probability that Y is less than 65.2 is the probability that a normal random variable Y comes out to be less than its mean by about 1.827 of Y's standard deviations, which is Phi(-1.827), the probability that a standard normal variable is less than -1.827, which is about 0.0339 . I find this value using the Excel formula

=NORM.S.DIST(-1.827, TRUE)

which evaluates to the cumulative distribution function of the standard normal cumulative distribution function at -1.827, which is defined as P(Z < -1.827) (where Z is standard normal, normal with mean 0 and sd 1.

So the probability that the average of a random sample of 16 independent pizza slices is less than 0.652 is about .0339 or 3.39%. ( By the way, round up here rather than down, to avoid exaggerating its already dramatic smallness.)

d)

Another very straightforward question.

We want to find (approximately) the 15 percentile of the random variable Y above. So for what value of a is P(Y < a) = 0.15? The 15 percentile of the standard normal distribution is computed in Excel as the value of the formula

=NORM.S.INV(0.15)

which is -1.03643 . So a is about 1.03643 sd's less than Y's mean.

Indeed, every normal random variable, like Y, has a 15th percentile that is 1.03643 sd's left (smaller than) its mean. So Y's 15 percentile is

a = 66.1 - 1.03643 x 0.4925 = 65.58956

or about 65.6 grams.

So about 15% of all random samples of 16 pizza slices will have average weights less than 65.6 grams.