Question:
The random variable \(X\) represents the number of girls in a family of three children. Assuming that boys and girls are equally likely, what are the mean and standard deviation for the random variable \(X\)?
The random variable \(X\) represents the number of girls in a family of three children. Assuming that boys and girls are equally likely, what are the mean and standard deviation for the random variable \(X\)?
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Recognize that \(X\) is Binomial with \(n = 3\), \(p = 0.5\); use \(E(X) = np\) and \(\text{Var}(X) = npq\).
Updated On: Jul 4, 2026
- Mean: 1.50; standard deviation: 0.87
- Mean: 1.50; standard deviation: 0.76
- Mean: 2.25; standard deviation: 0.87
- Mean: 2.25; standard deviation: 0.76
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The Correct Option is A
Solution and Explanation
Step 1: Identify the distribution. Each child is independently a girl with probability \(p = 0.5\) or a boy with probability \(q = 1 - p = 0.5\). With \(n = 3\) children, the number of girls \(X\) follows a Binomial distribution with parameters \(n = 3\) and \(p = 0.5\), so \(X \sim B(3, 0.5)\).
Step 2: Compute the mean. For a Binomial distribution, \[E(X) = np = 3 \times 0.5 = 1.50\]
Step 3: Compute the variance. \[\text{Var}(X) = npq = 3 \times 0.5 \times 0.5 = 0.75\]
Step 4: Compute the standard deviation. \[\text{SD}(X) = \sqrt{0.75} \approx 0.87\]
Step 5: So Mean = 1.50 and SD = 0.87. \[\boxed{\text{Mean } 1.50,\ \text{SD } 0.87}\]
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