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Math Help - Normal Distribution

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    Normal Distribution

    I have a statistics problem I canít solve for some reason:

    A blackjack player at a Las Vegas casino learned that the house will provide a free room if play is for four hours at an average bet of $50. The playerís strategy provides a probability of .49 of winning on any one hand, and the player knows that there are 60 hands per hour. Suppose the player plays for four hours at a bet of $50 per hand.

    b. Whatís the probability the player loses $1000 or less? Answer: .1788

    For some reason I canít get their answer. It looks like a really simple normal approximation of binomial probabilities question, but for some reason none of my answers turn out right.

    I got the expected value of this game as -240, because he wages $12,000 total and the difference between his expected winnings and expected losses is -240.

    This is the part where I donít know how I went wrong: I found standard deviation using sqrt(12000*.49*(1-.49)), which is from sqrt(n*p*(1-p)). Itís 54.7613

    Next, I find the z-score, which comes out as -13.8689. I got this using z=(x-u)/s

    -13.8689 is so high of a z value itís not even on my chart. What did I do wrong?
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    Quote Originally Posted by lisakki View Post
    I have a statistics problem I can’t solve for some reason:

    A blackjack player at a Las Vegas casino learned that the house will provide a free room if play is for four hours at an average bet of $50. The player’s strategy provides a probability of .49 of winning on any one hand, and the player knows that there are 60 hands per hour. Suppose the player plays for four hours at a bet of $50 per hand.

    b. What’s the probability the player loses $1000 or less? Answer: .1788

    For some reason I can’t get their answer. It looks like a really simple normal approximation of binomial probabilities question, but for some reason none of my answers turn out right.

    I got the expected value of this game as -240, because he wages $12,000 total and the difference between his expected winnings and expected losses is -240.

    This is the part where I don’t know how I went wrong: I found standard deviation using sqrt(12000*.49*(1-.49)), which is from sqrt(n*p*(1-p)). It’s 54.7613

    Next, I find the z-score, which comes out as -13.8689. I got this using z=(x-u)/s

    -13.8689 is so high of a z value it’s not even on my chart. What did I do wrong?
    Let X be the random variable number of games won.

    X ~ Binomial(n = 240, p = 0.49)

    If the player loses $1,000 then X = 110. So you need to calculate \Pr(X \leq 110) using the normal approximation.
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