In the game of roulette a wheel consists of 18 slots numbered 0,00,1, 2..., 36 t

Question

In the game of roulette a wheel consists of 18 slots numbered 0,00,1, 2..., 36 to play the game a metal bail is spun around re wheel and is allowed to tall into one of the numbered Mots lithe number of the lot the ball falls into matches the number you selected, you win $35 otherwise you lose $1 Complete parts (a) through (g) below Suppose that you play He game 90 times that n = 90 the sampling mean amount won per game What are Bio moan and standard deviation of the sampling Round your results to me noarosl penny What is the probability of being ahead after playing the game 90 times? That inability that mean it greater ran 0 tot n = 90?

Explanation / Answer

a)

There are 38 numbers, so winning probability is 1/38, and losing at 37/38.

b)

Consider:

Thus,  
  
u = Expected value = mean = Sum(xP(x)) = -0.052631579 [ANSWER]

Var(x) = E(x^2) - E(x)^2 =    33.20775623

sigma(x) = sqrt [Var(x)] =    5.762617134 [ANSWER]

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c)

By central limit theorem it will be approximately normally distirbuted.

It has the same mean,

u(x) = -0.052631579 [ANSWER]

and a reduced standard deviation,

sigma(x) = sigma/sqrt(n) = 5.762617134/sqrt(90) = 0.607433181 [ANSWER]

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d)

Being ahead means a mean greater than 0.

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    0      
u = mean =    -0.052631579      
          
s = standard deviation =    0.607433181      
          
Thus,          
          
z = (x - u) / s =    0.086645874      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.086645874   ) =    0.4654765 [ANSWER]

x P(x) 35 0.026316 -1 0.973684

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