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There were 49.7 million people with some type of long-lasting condition or disability living in the United States in 2000. This represented 19.3 percent of the majority of civilians aged five and over (http://factfinder.census.gov). A sample of 1000 persons is selected at random.
a. Approximate the probability that more than 200 persons in the sample have a disability.
b. Approximate the probability that between 180 and 300 people in the sample have a disability.

Sagot :

Answer:

The answer is "0.274 and 0.841"

Step-by-step explanation:

Given:

In this question, let X  the binomial random variable with the parameters:

[tex]n = 1000\\\\P = 0.193[/tex]

Calculating the random variable Z:

[tex]Z= \frac{X - np}{\sqrt{np(l - p)}}=\frac{x -193}{12.48}[/tex]

is an approximately regular, cumulative random variable with [tex]\Phi[/tex] specified in the back of the book in the tables.

Using the continuity and tables to calculate:

[tex]\to \mathbb{P} (X > 200) = \mathbb{P} (X\geq 200.5) \approx \mathbb{P}(Z\geq 0.6)=1- \Phi(0.6) \approx 0.274 \\\\\to \mathbb{P} (180 < X < 300) = \mathbb{P} (180.5 \leq X \leq 299.5) = \approx \mathbb{P} (-1 \leq Z\leq 8.53) = \approx 1- \Phi(-1) \approx 0.841[/tex]

Cricetus

(a) The probability that more than 200 persons throughout the sample have a disability will be "0.08691".

(b) The probability that between 180 - 300 people throughout the sample have a disability will be "0.85122".

Probability

According to the question,

Majority percentage, P = 19.3% or,

                                       = 0.193

Sample person, n = 1000

Q = 1 - P

   = 1 - 0.193

   = 0.807

By using Normal approx. to Binomial,

→ Mean, E(X) = n × P

By substituting the values,

                     = 1000 × 0.193

                     = 193

Standard deviation, X = √n × p × Q

                                      = √1000 × 0.193 × 807

                                      = 12.4800  

(a)

Z = [tex]\frac{X - \mu}{\sigma}[/tex]

  = [tex]\frac{210-193}{12.48}[/tex]

  = 1.36

The probability be:

→ P(X > 210) = 1 - P (Z < 1.36)

                    = 1 - 0.9131

                    = 0.08691

(b)

Z = [tex]\frac{X - \mu}{\sigma}[/tex]

  = [tex]\frac{180-193}{12.48}[/tex] or, [tex]\frac{300-193}{12.48}[/tex]

  = 8.57

hence,

The probability be:

→ P (180 < X < 300) = P(Z < 8.57) - P(Z < -1.04)

                                = 1 - 0.1488

                                = 0.85122  

Thus the answers above are correct.

Find out more information about probability here:

https://brainly.com/question/24756209

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