Math Problem Statement

According to a certain organization, approximately 26% of adults in the United States live with a disability. This includes disabilities affecting mobility, cognition, vision, hearing, and others. Suppose that we select a random sample of five US adults, and let x = the number of adults out of five that live with a disability. The methods of Section 6.2 can be used to compute the probability assignments for the x distribution. Find the probabilities for parts (a) through (e).

Solution

Let's work through each part of this problem based on the probability distribution table provided:

(a) Probability that one or more adults live with a disability:

The probability that none of the five adults live with a disability is given by P(x=0)=0.222P(x = 0) = 0.222. The probability that one or more adults live with a disability is the complement of this event:

P(one or more)=1P(x=0)P(\text{one or more}) = 1 - P(x = 0) P(one or more)=10.222=0.778P(\text{one or more}) = 1 - 0.222 = 0.778

(b) Probability that two or more adults live with a disability:

The probability that two or more adults live with a disability can be found by subtracting the probabilities for zero and one adult living with a disability from 1:

P(two or more)=1(P(x=0)+P(x=1))P(\text{two or more}) = 1 - (P(x = 0) + P(x = 1)) P(two or more)=1(0.222+0.390)=10.612=0.388P(\text{two or more}) = 1 - (0.222 + 0.390) = 1 - 0.612 = 0.388

(c) Probability that three or more adults live with a disability:

Similarly, the probability that three or more adults live with a disability can be found by subtracting the probabilities for zero, one, and two adults from 1:

P(three or more)=1(P(x=0)+P(x=1)+P(x=2))P(\text{three or more}) = 1 - (P(x = 0) + P(x = 1) + P(x = 2)) P(three or more)=1(0.222+0.390+0.274)=10.886=0.114P(\text{three or more}) = 1 - (0.222 + 0.390 + 0.274) = 1 - 0.886 = 0.114

(d) Expected number of adults with a disability:

The expected value μ\mu is calculated using the formula for the mean of a probability distribution:

μ=[xP(x)]\mu = \sum [x \cdot P(x)]

Calculating step by step:

μ=(0×0.222)+(1×0.390)+(2×0.274)+(3×0.096)+(4×0.017)+(5×0.001)\mu = (0 \times 0.222) + (1 \times 0.390) + (2 \times 0.274) + (3 \times 0.096) + (4 \times 0.017) + (5 \times 0.001) μ=0+0.390+0.548+0.288+0.068+0.005=1.299\mu = 0 + 0.390 + 0.548 + 0.288 + 0.068 + 0.005 = 1.299

(e) Standard deviation of adults with a disability:

The standard deviation σ\sigma is calculated using the formula:

σ=[x2P(x)]μ2\sigma = \sqrt{\sum [x^2 \cdot P(x)] - \mu^2}

First, compute [x2P(x)]\sum [x^2 \cdot P(x)]:

[x2P(x)]=(02×0.222)+(12×0.390)+(22×0.274)+(32×0.096)+(42×0.017)+(52×0.001)\sum [x^2 \cdot P(x)] = (0^2 \times 0.222) + (1^2 \times 0.390) + (2^2 \times 0.274) + (3^2 \times 0.096) + (4^2 \times 0.017) + (5^2 \times 0.001) [x2P(x)]=0+0.390+1.096+0.864+0.272+0.025=2.647\sum [x^2 \cdot P(x)] = 0 + 0.390 + 1.096 + 0.864 + 0.272 + 0.025 = 2.647

Now, calculate the standard deviation:

σ=2.647(1.299)2=2.6471.687=0.9600.98\sigma = \sqrt{2.647 - (1.299)^2} = \sqrt{2.647 - 1.687} = \sqrt{0.960} \approx 0.98

Final Answers:

  • (a) Probability one or more: 0.778
  • (b) Probability two or more: 0.388
  • (c) Probability three or more: 0.114
  • (d) Expected number μ\mu: 1.30
  • (e) Standard deviation σ\sigma: 0.98

Would you like further details or have any other questions?

Here are 5 related questions to explore:

  1. What is the probability that exactly 2 adults have a disability?
  2. How would the expected value change if the sample size increased to 10 adults?
  3. What does the complement rule in probability represent, and how is it used here?
  4. How is standard deviation interpreted in this context?
  5. How could a change in the overall percentage of adults with a disability (from 26%) affect this distribution?

Tip: Always check that the total sum of probabilities in a probability distribution equals 1.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution
Expected Value
Standard Deviation

Formulas

Complement Rule: P(one or more) = 1 - P(none)
P(two or more) = 1 - (P(x = 0) + P(x = 1))
P(three or more) = 1 - (P(x = 0) + P(x = 1) + P(x = 2))
Expected value: μ = Σ[x * P(x)]
Standard deviation: σ = √[Σ(x^2 * P(x)) - μ^2]

Theorems

Law of Total Probability
Properties of Expected Value
Properties of Variance and Standard Deviation

Suitable Grade Level

College / University Level