# I A level S1&S2 Bionomial help

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1. Aug 24, 2016

Getting ahead with my S1 & S2 for my further maths mods next year, really stuck on this particular question, as there is no such successive trials etc, please help me solve, so i can make some actual sense of what the hell is going on! Cheers,

roughly that the portion of families with
access to clean drinking water is equal to 0.55. Find the probability that of a
sample n, over 40% will have clean drinking water i n=6 binomial
distribution, and ii n=600 Gaussian distribution

2. Aug 24, 2016

### Lucas SV

Set $p=0.55$ to be the probability that a randomly chosen family will have access to clean water. For the first case you are asking the probability that 3, 4, 5 or 6 families will have access to clean water (that is over 40% of 6). So you essentially sum the probalities $P(3)$ to $P(6)$, using thee formula for binomial distribution (probability mass function).

For the second case you use the cummulative distribution function for the gaussian distribution. The justification is the central limiting theorem, which allows you to approximate the binomial distribution by the normal distribution for large $n$. You will need to calculate the mean and variance (formulas given by the binomial distribution). You may be given a table (you can use a graphical calculator or wolfram alpha) that allows you to compute the CDF. So in your case you need to compute $1-CDF(240)$, which is the probability that over 40% of the 600 families will have access to clean water.

Last edited: Aug 24, 2016
3. Aug 24, 2016

Hi lucas, thanks for your quick response, if possible please could show me the solution to the second the problem? Im new to gaussian, watched a few youtube tutorials, but i've been on this problem for a few hours now and just need to see how it's broken down and solved. Thanks again for your help Lucas!

4. Aug 24, 2016

### Lucas SV

Sure! The mean is $np=600*0.55=330$ and the standard deviation is $np(1-p)=600*0.55*0.45=148.5$. Then according to http://www.danielsoper.com/statcalc/calculator.aspx?id=53, with $x=240=0.4*600$, I find essentially find the area under the normal distribution function from $-\infty$ to $x$, to be $0.2722$. The answer then is $1-0.2722=0.728$ (to 3 significant figures), which is the area under the graph from 240 to $+\infty$. You can check the result in a table or a different calculator and it will be the same.

Note that the central limit theorem tells us that this is a good approximation for large enough $n$ and for $x$ which is sufficiently close to the mean. The reason why you approximate is because it requires a lot of computational power to compute probabilities exactly (the way you did in part a) for large $n$ in a binomial distribution (which describes the probability of $k$ successes in $n$ trials).

5. Aug 24, 2016