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 !
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:
$\large{P(X\leq x) \approx 0.5 \cdot \Bigg( 1 + \sqrt{1-e^{- (\sqrt{\frac{\pi}{8}} \cdot x^2)}} \Bigg)}$
Normalverteilung (german)=normal distribution