# request for normal distribution exercise

• August 4th 2010, 08:00 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, 04:59 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 4th 2010, 11:41 PM
astipd

I figured out a solution. Do you think this is correct?

p=0,5
X ~ Bin(100, 0.5)

X is approximately normally distributed with μ = (n × p) = 100 × 0.5 = 50 and
σ2 = n × p × (1-p) = 100 × 0.5 × 0.5 = 25
X is approximately N(50,25)

P (x≥50) = 1 – P(x < 50) = 1 – P (z<(50-μ)/σ) = 1 - P (z<(50-50)/5) = 1 - P(z<0) = 1 – Φ(0) = 0.5