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Residences of U.S. Citizens According to the U.S. Census, 67.5% of the U.S. population were bom in their state of residence. In a random sample of 222 Americans, what is the probability that fewer than 136 were born in their state of residence? Round the final answer to at least four decimal places and intermediate z-value calculations to two decimal places. P(x<136)=

User Antiohia
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Final answer:

The question is about finding the probability of a binomial proportion, and it is solved using a normal approximation to find the probability of fewer than 136 people in the sample being born in their state of residence.

Step-by-step explanation:

The student is asking for the probability that fewer than 136 people in a random sample of 222 were born in their state of residence, given that 67.5% of the U.S. population were born in their state of residence. This is a binomial probability problem that can be approximated using the normal distribution since the sample size is large enough.

To find the probability P(x < 136), we first need to calculate the mean (μ) and standard deviation (σ) of the binomial distribution:

  • μ = n * p = 222 * 0.675
  • σ = √(n * p * (1 - p)) = √(222 * 0.675 * (1 - 0.675))

Then we calculate the z-score for 135.5 (we use 135.5 instead of 136 due to continuity correction) using the formula:

z = (x - μ) / σ

Finally, we look up this z-score in the standard normal distribution table, or use a calculator, to find the probability that corresponds to this z-score. The calculated probability will be the answer to the student's question

User Jpmc
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