Subjects probability

Coin Flip Variance 73B6F0

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Coin Flip Variance 73B6F0


1. **Problem Statement:** We flip four coins and define the random variable $X$ as the number of heads in the four tosses. We need to calculate the variance and standard deviation of $X$ after listing the sample space, possible values, type of variable, and probability distribution. 2. **Sample Space and Values:** The sample space consists of all sequences of H (heads) and T (tails) for four tosses, totaling $2^4=16$ outcomes. 3. **Possible values of $X$:** Since $X$ counts heads, possible values are $0,1,2,3,4$. 4. **Probability Distribution:** The probability of getting exactly $k$ heads in 4 tosses follows a binomial distribution: $$P(X=k) = \binom{4}{k} \left(\frac{1}{2}\right)^4 = \binom{4}{k} \frac{1}{16}$$ Calculating each: - $P(0) = \binom{4}{0} \frac{1}{16} = 1 \times \frac{1}{16} = \frac{1}{16}$ - $P(1) = \binom{4}{1} \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16}$ - $P(2) = \binom{4}{2} \frac{1}{16} = 6 \times \frac{1}{16} = \frac{6}{16}$ - $P(3) = \binom{4}{3} \frac{1}{16} = 4 \times \frac{1}{16} = \frac{4}{16}$ - $P(4) = \binom{4}{4} \frac{1}{16} = 1 \times \frac{1}{16} = \frac{1}{16}$ 5. **Calculate the Expected Value $E(X)$:** $$E(X) = \sum_{k=0}^4 k P(X=k) = 0 \times \frac{1}{16} + 1 \times \frac{4}{16} + 2 \times \frac{6}{16} + 3 \times \frac{4}{16} + 4 \times \frac{1}{16}$$ $$= 0 + \frac{4}{16} + \frac{12}{16} + \frac{12}{16} + \frac{4}{16} = \frac{32}{16} = 2$$ 6. **Calculate the Variance $Var(X)$:** Variance formula: $$Var(X) = E(X^2) - [E(X)]^2$$ First find $E(X^2)$: $$E(X^2) = \sum_{k=0}^4 k^2 P(X=k) = 0^2 \times \frac{1}{16} + 1^2 \times \frac{4}{16} + 2^2 \times \frac{6}{16} + 3^2 \times \frac{4}{16} + 4^2 \times \frac{1}{16}$$ $$= 0 + \frac{4}{16} + \frac{24}{16} + \frac{36}{16} + \frac{16}{16} = \frac{80}{16} = 5$$ Now calculate variance: $$Var(X) = 5 - (2)^2 = 5 - 4 = 1$$ 7. **Calculate the Standard Deviation $\sigma$:** Standard deviation is the square root of variance: $$\sigma = \sqrt{Var(X)} = \sqrt{1} = 1$$ **Final answers:** - Variance $Var(X) = 1$ - Standard Deviation $\sigma = 1$