a. To find the probability of a household having 6 or more people, we need to sum up the probabilities for a household of size 6 or more.
P(6 or more) = Probability of a household being of size 6 or more = ?
Given data:
Size of Household: 1, 2, 3, 4, 5, 6 or more
Probabilities: 10.8%, 31.3%, 12.4%, 28.7%, 11.6%, ?
To find P(6 or more):
P(6 or more) = 100% - (P(1) + P(2) + P(3) + P(4) + P(5))
P(6 or more) = 100% - (10.8% + 31.3% + 12.4% + 28.7% + 11.6%)
P(6 or more) = 100% - 94.8%
P(6 or more) = 5.2%
Therefore, the probability of a household having 6 or more people is 5.2%.
b. To find the expected value of household sizes in the U.S., we multiply each household size by its corresponding probability and sum them up.
Expected value (mean) = E(X) = Σ (x * P(x))
Using the given data and assuming 6 or more as size 6:
E(X) = (1 * 0.108) + (2 * 0.313) + (3 * 0.124) + (4 * 0.287) + (5 * 0.116) + (6 * 0.052)
E(X) = 0.108 + 0.626 + 0.372 + 1.148 + 0.58 + 0.312
E(X) = 3.146
Therefore, the expected value of household sizes in the U.S. is 3.146.
c. To find the standard deviation of household sizes, we will use the formula:
Standard deviation = sqrt[ Σ [ (x - E(X))^2 * P(x) ] ]
Substitute the values calculated in part b:
Standard deviation = sqrt[ ( (1-3.146)^2 * 0.108 ) + ( (2-3.146)^2 * 0.313 ) + ( (3-3.146)^2 * 0.124 ) + ( (4-3.146)^2 * 0.287 ) + ( (5-3.146)^2 * 0.116 ) + ( (6-3.146)^2 * 0.052 ) ]
Standard deviation = sqrt[ (2.146^2 * 0.108) + (1.146^2 * 0.313) + (0.146^2 * 0.124) + (0.854^2 * 0.287) + (1.854^2 * 0.116) + (2.854^2 * 0.052) ]
Standard deviation = sqrt[ 0.495 + 0.423 + 0.002 + 0.329 + 0.378 + 0.41 ]
Standard deviation ≈ sqrt[ 2.037 ]
Standard deviation ≈ 1.428
Therefore, the standard deviation of household sizes is approximately 1.428.
\textbf{Answer:}
a. The probability of a household having 6 or more people is 5.2%.
b. The expected value of household sizes in the U.S. is 3.146.
c. The standard deviation of household sizes is approximately 1.428.