Binomial normal distribution (1 Viewer)

NexusRich · Oct 26, 2021

Can someone please provide working out and answers for these 2 questions, the answers that I got is totally different to that of the textbook and I dont really know what went wrong.

5uckerberg · Oct 26, 2021

NexusRich said:
Can someone please provide working out and answers for these 2 questions, the answers that I got is totally different to that of the textbook and I dont really know what went wrong.

Note that the $E(X)=np=50\times{0.5}=25$ and $SD(X)=\sqrt{12.5}=\frac{5}{\sqrt{2}}$ , and the next step will be interesting. This is between the first and second standard deviations and the third deviations. There use the infamous z-table to your advantage.

In general these type of questions leverage the use of $E(X)=np, SD(X)=\sqrt{np(1-p)}$

Apologies for my rushed explanation. Actually one can do this in an easier fashion. Here is the routine. To find the z-score range this formula is used $z=\frac{X-\mu}{\sigma}, \sigma=SD(X), \mu=E(X)$ . Now sub the range of X okay and see the range of z-scores involved. There you can find the probability.

5uckerberg · Oct 26, 2021

Note to all current and future Mathematics Extension 1 candidates in Years 11 and 12. Please note that for the binomial normal distribution there is an interesting pattern $E(X)=np, Var(X)=np(1-p), SD(X)=\sqrt{np(1-p)}.$ Why is that the case?

Well, the big important aspect in this question goes like this. If one recalls what happens when they found the expected value for continuous probability distributions they always had to go with the sigma notation times the probability because there were many different outcomes involved and through that, you can find the expected value. However, in Extension I mathematics because there is only one result that we focus on and one probability and as such you have $E(X)=np$ instead of $E(X)=n_{1}p_{1}+n_{2}p_{2}+n_{3}p_{3}+...$ which caters to multiple outcomes and a variety of chances in terms of outcomes. Now $Var(X)=E(X^{2})-(E(X)^{2})$ . Now the variance is this

My question to you is how do we make use of this in Extension I mathematics?

Okay, the idea is very clear so here is what happens next, the definition for the variance if one expands the function is just simply $Var(X)=E(X^{2})-(E(X)^{2})$ . Now we can apply this idea here where what we will have is $\frac{n^{2}p-(np)^{2}}{n}=np(1-p).$ Noting that n is the number of results and prefers to the probability of results. Then the standard deviation is this $\sqrt{np(1-p)}.$ Note the similiarity to the Mathematics Advanced version for standard deviation, to which we call $\omega.$

Binomial normal distribution (1 Viewer)

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