Questions: In the probability distribution to the right, the random variable X represents the number of marriages an individual aged 15 years or older has been involved in. Complete parts (a) through (f) below. x P(x) 0 0.261 1 0.584 2 0.122 3 0.028 4 0.004 5 0.001 B. If any number of individuals aged 15 year or older were surveyed, the sample mean number of marriages should be close to the mean of the random variable. C. If many randomly selected individuals 40 to 49 years of age were surveyed, the sample mean number of marriages should be close to the mean of the random variable. D. If many individuals aged 15 year or older were surveyed, the sample mean number of marriages should be close to the mean of the random variable. E. If many randomly selected individuals aged 15 years or older were surveyed, the observed number of marriages will be equal to the mean number of marriages for most individuals. (d) Compute the standard deviation of the random variable X . σX=0.74 marriages (Round to two decimal places as needed.) (e) What is the probability that a randomly selected individual 15 years or older was involved in two marriages? (Type an integer or a decimal. Do not round.)

In the probability distribution to the right, the random variable X represents the number of marriages an individual aged 15 years or older has been involved in. Complete parts (a) through (f) below. x P(x) 0 0.261 1 0.584 2 0.122 3 0.028 4 0.004 5 0.001 B. If any number of individuals aged 15 year or older were surveyed, the sample mean number of marriages should be close to the mean of the random variable. C. If many randomly selected individuals 40 to 49 years of age were surveyed, the sample mean number of marriages should be close to the mean of the random variable. D. If many individuals aged 15 year or older were surveyed, the sample mean number of marriages should be close to the mean of the random variable. E. If many randomly selected individuals aged 15 years or older were surveyed, the observed number of marriages will be equal to the mean number of marriages for most individuals. (d) Compute the standard deviation of the random variable X . σX=0.74 marriages (Round to two decimal places as needed.) (e) What is the probability that a randomly selected individual 15 years or older was involved in two marriages? (Type an integer or a decimal. Do not round.)
Transcript text: In the probability distribution to the right, the random variable X represents the number of marriages an individual aged 15 years or older has been involved in. Complete parts (a) through ( $f$ ) below. \begin{tabular}{ll} $x$ & $P(x)$ \\ \hline 0 & 0.261 \\ 1 & 0.584 \\ 2 & 0.122 \\ 3 & 0.028 \\ 4 & 0.004 \\ 5 & 0.001 \\ \hline \end{tabular} B. If any number of individuals aged 15 year or older were surveyed, the sample mean number of marriages should be close to the mean of the random variable. C. If many randomly selected individuals 40 to 49 years of age were surveyed, the sample mean number of marriages should be close to the mean of the random variable. D. If many individuals aged 15 year or older were surveyed, the sample mean number of marriages should be close to the mean of the random variable. E. If many randomly selected individuals aged 15 years or older were surveyed, the observed number of marriages will be equal to the mean number of marriages for most individuals. (d) Compute the standard deviation of the random variable X . $\sigma_{\mathrm{X}}=0.74$ marriages (Round to two decimal places as needed.) (e) What is the probability that a randomly selected individual 15 years or older was involved in two marriages? $\square$ (Type an integer or a decimal. Do not round.)
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Solution

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Solution Steps

Step 1: Calculate the Mean

The mean \( \mu \) of the random variable \( X \) is calculated using the formula:

\[ \mu = \sum_{x} x \cdot P(x) = 0 \times 0.261 + 1 \times 0.584 + 2 \times 0.122 + 3 \times 0.028 + 4 \times 0.004 + 5 \times 0.001 = 0.933 \]

Step 2: Calculate the Variance

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \sum_{x} (x - \mu)^2 \cdot P(x) = (0 - 0.933)^2 \times 0.261 + (1 - 0.933)^2 \times 0.584 + (2 - 0.933)^2 \times 0.122 + (3 - 0.933)^2 \times 0.028 + (4 - 0.933)^2 \times 0.004 + (5 - 0.933)^2 \times 0.001 = 0.543 \]

Step 3: Calculate the Standard Deviation

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{0.543} = 0.737 \]

Step 4: Calculate the Probability of Two Marriages

The probability that a randomly selected individual aged 15 years or older was involved in two marriages is given by:

\[ P(X = 2) = 0.122 \]

Final Answer

The results are summarized as follows:

  • Mean: \( \mu = 0.933 \)
  • Variance: \( \sigma^2 = 0.543 \)
  • Standard Deviation: \( \sigma = 0.737 \)
  • Probability of being involved in two marriages: \( P(X = 2) = 0.122 \)

Thus, the final answers are: \[ \boxed{\mu = 0.933} \] \[ \boxed{\sigma^2 = 0.543} \] \[ \boxed{\sigma = 0.737} \] \[ \boxed{P(X = 2) = 0.122} \]

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