# request for normal distribution exercise

• August 4th 2010, 09:00 AM
astipd
request for normal distribution exercise

The probability to win in a roulette game is 18/37. If someone bets 100 times, which is the probability to win at least 50 times?
(an excellent approximation to B(n, p) is given by the normal distribution, where μ=np and σ2=np(1-p))

Solution

P (50 ≤ x ≤ 100) = P ((50-μ)/σ ≤( x-μ )/σ≤( 100-μ)/σ) = Φ((100-μ)/σ) – Φ((50-μ)/σ)

[ where μ=np, σ2=np(1-p), n=100, p=18/37 ]

then we use the normal distribution table
• August 4th 2010, 05:59 PM
pickslides
You are on the right track. Consider where $X\text{~ Bi}(n,p)$ and $Y\text{~ N}(\mu,\sigma)$ then $P(X\leq x) \approx P(Y\leq x+0.5)$
• August 5th 2010, 12:41 AM
astipd