Here is the question I need help on :

Let $X$ be a binomial random variable with p = 0.5 and n = 100.

Give $P(X \geq 60)$ rounded to two decimal places without using a calculator (by using approximations I assume).

I easily find that $ \displaystyle P(X \geq 60) = 1 - \sum_{k=0}^{59} \binom {100} {k}(0.5)^{100}$ or alternatively $\displaystyle \sum_{k=60}^{100} \binom {100}{k} (0.5)^{100}$. But I can't seem to find a good approximation of this that doesn't require the use of a calculator or a little program.

Thanks for the help !

  • 1
    $\begingroup$ I would guess they are expecting you to use a normal approximation. $\endgroup$ – André Nicolas Nov 6 '14 at 17:37
  • $\begingroup$ That leads to be able to approximate the cumulative distribution function of a normal variable, any clue? $\endgroup$ – Raven Nov 6 '14 at 17:41
  • $\begingroup$ you can use the z-score and then use the tabulated values to generate the required probability. $\endgroup$ – Chinny84 Nov 6 '14 at 17:43

You can make an approximation by the normal distribution:

$\large{1-P(X\leq 59)\approx 1-\Phi \left( \frac{59+0.5-50}{\sqrt{100\cdot 0.5 \cdot 0.5}} \right)}$

$\Phi(z)$ is the cummulative function of the standard normal distribution.

0.5 is the continuous correction factor.

Formula for approximation the cdf of the standard normal distribution:

$\large{P(X\leq x) \approx 0.5 \cdot \Bigg( 1 + \sqrt{1-e^{- (\sqrt{\frac{\pi}{8}} \cdot x^2)}} \Bigg)}$

enter image description here

Normalverteilung (german)=normal distribution

  • $\begingroup$ Thanks. I came up with this but I couldn't approximate $\Phi(9.5/5)$ without a table, is there a simple way to do it ? Or can I approximate it by $\Phi(2)$ (I mean, am I supposed to know $\Phi(2)$ by heart ?) $\endgroup$ – Raven Nov 6 '14 at 18:01
  • $\begingroup$ Nevermind, $\Phi(1.96)$ = 0.975. Thanks a lot. $\endgroup$ – Raven Nov 6 '14 at 18:15
  • $\begingroup$ Just wait. In general you need a table. But long ago I have found a formula, which approximate the standard normal distribution. The picture visualizes it. It is your decision, if it is exactly enough for you. I have made an edit. $\endgroup$ – callculus Nov 6 '14 at 18:29
  • $\begingroup$ I've seen the formula, thanks that's a great visualisation. But, without a calculator, it's impossible to calculate the value of this $\endgroup$ – Raven Nov 6 '14 at 19:15
  • $\begingroup$ Yes, without a calculator you are lost. You can learn the standard values of z (1.96, 2.33, ...) by heart and hope that this what you need. But this is not really a sure road. The posted formula is the best simplification I know. $\endgroup$ – callculus Nov 6 '14 at 19:27

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