binomial distribution problemo

Question

A fair coin is tossed n times. The histogram for the resulting binomial distribution is labelled 0, 1, 2, . . . , n on the horizontal axis, and each column is 1 unit wide. How many columns are entirely contained in the interval one standard deviation or less from the mean, when n is 16?

working...

Let the point on the Binomial Distribution histogram that is one standard deviation less than the mean be called lower bound "a" and the point one standard deviation above the mean be called upper bound "b". And they tell us the width of the columns is 1. So I use the binomial distribution expressions for mean and the standard deviation given that p = 0.5 ( we are tossing a coin) to eventually get that the interval in between "a" and "b" is exactly $\sqrt n$. Hence th number of columns that fit is $\sqrt n$ divided by 1 which is $\sqrt n $ = k columns that fit. I arrived at the answer that the number of columns that fit within that range is k=$\sqrt n$ which for n= 16 yields k=4 so 4 columns can fit.

I'm not sure if my answer is correct because it seems to lie on the assumption that, in a BD histogram, the columns start exactly on the lower bound "a" up to b. Please see the photo below to understand.

Answer 1

Okay, so any histogram for a binomial distribution is going to be counting integer results - means aren't always integers, of course. We don't necessarily have to assume that the columns start exactly at $a$ units, but a reasonable assumption is instead that columns always start on an integer unit mark, and are exactly 1 unit wide. So, the column for 0 successes starts at 0 and ends at 1, the column for 1 success starts at 1 and ends at two, etc.

So, that decouples our columns from $a$ and $b$. We still want to know how many whole columns fit between $a$ and $b$. With the assumption above, that means that we want to calculate $\lceil b\rceil-\lfloor a\rfloor$ - aka, the integer component of $b$ minus the next integer after $a$.

Let's look at your example and calculate with this in mind:

\begin{align*} \sigma&=\sqrt{np(1-p)}&\mu=&np\\ &=\sqrt{16(0.5)(0.5)}& =&16(0.5)\\ &=\sqrt{4} & =&8 \\ &=2\\ \\ a&= \mu - \sigma &b=&\mu + \sigma\\ &=8-2&=&8+2\\ &=6&=&10\\ \\ C&=\lfloor b \rfloor-\lceil a \rceil\\ &=10-6\\ &=4 \end{align*}

So, your answer still holds for your particular example - there will be exactly 4 columns of space between $a$ and $b$, but this example is a best case - both $\mu$ and $\sigma$ are integers, so there's absolutely no lost space. (although, it's worth noting that because the mean is an integer, it's taking up one of those 4 columns, so the histogram will be lopsided - 1 column below the mean, the mean, and 2 columns above the mean).

Let's try this with $p=0.6$ instead and see what we get:

\begin{align*} \sigma&=\sqrt{np(1-p)}&\mu=&np\\ &=\sqrt{16(0.6)(0.4)}& =&16(0.4)\\ &=\sqrt{3.84} & =&9.6 \\ &\approx 1.9595917942\\ \\ a&= \mu - \sigma &b=&\mu + \sigma\\ &= 9.6-\sqrt{3.84}&=&9.6+\sqrt{3.84}\\ &\approx 7.64040821&\approx& 11.55959179\\ \\ C&=\lfloor b \rfloor-\lceil a \rceil\\ &=11-8\\ &=3 \end{align*}

So, this gives us a lower answer, although we can see that there's quite a bit of wasted space - in this case, $b-a \approx 3.91918358$, so there's nearly a unit of wasted space that cannot hold an entire column.