According to previous studies, 16% of the U.S. population is left-handed. Not knowing this, a high school student claims that the percentage of left-handed people in the U.S. is 15%.The student is going to take a random sample of 2100 people in the U.S. to try to gather evidence to support the claim. Let p be the proportion of left-handed people in the sample.Answer the following. (If necessary, consult a list of formulas.)(a)Find the mean of p.(b)Find the standard deviation of p.(c)Compute an approximation for P≥p0.15, which is the probability that there will be 15% or more left-handed people in the sample. Round your answer to four decimal places.
Question
According to previous studies, 16% of the U.S. population is left-handed. Not knowing this, a high school student claims that the percentage of left-handed people in the U.S. is 15%.The student is going to take a random sample of 2100 people in the U.S. to try to gather evidence to support the claim. Let p be the proportion of left-handed people in the sample.Answer the following. (If necessary, consult a list of formulas.)(a)Find the mean of p.(b)Find the standard deviation of p.(c)Compute an approximation for P≥p0.15, which is the probability that there will be 15% or more left-handed people in the sample. Round your answer to four decimal places.
Solution
(a) The mean of p is equal to the population proportion, which is 0.16 or 16%.
(b) The standard deviation of p can be calculated using the formula for the standard deviation of a sample proportion: sqrt[(P(1 - P)) / n], where P is the population proportion and n is the sample size. Substituting the given values, we get sqrt[(0.16 * 0.84) / 2100] = 0.0085.
(c) To compute an approximation for P≥0.15, we first need to find the z-score for 0.15. The z-score is calculated as (p - P) / standard deviation. Substituting the given values, we get (0.15 - 0.16) / 0.0085 = -1.1765. We then look up this z-score in a standard normal distribution table or use a calculator with a normal distribution function to find the probability. The probability for a z-score of -1.1765 is approximately 0.1200. However, since we are looking for the probability that p is greater than or equal to 0.15, we need to subtract this value from 1 (because the total probability under the normal curve is 1). So, 1 - 0.1200 = 0.8800. Therefore, the probability that there will be 15% or more left-handed people in the sample is approximately 0.8800 or 88.00%.
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