A survey conducted in 2011 asked employed adults from age 25 to 70 if they plan to keep a foot in the working world after retirement. Thirty-three percent of the respondents answered "very likely". A random sample of 75 adults from this age group was asked the same question. Using the normal approximation to the binomial distribution, what is the probability that more than 18 adults from this sample would very likely plan to keep a foot in the working world after retirement?

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To find the probability that more than 18 adults from the sample of 75 would very likely plan to keep a foot in the working world after retirement, we use the normal approximation to the binomial distribution. The mean (μ) of the distribution is given by n*p, where n is the sample size (75) and p is the probability of success (0.33, since 33% of the respondents answered "very likely"). Thus, μ = 75 * 0.33 = 24.75. The standard deviation (σ) is given by sqrt(n*p*(1-p)), which equals sqrt(75*0.33*(1-0.33)) ≈ 4.27.To find the probability of more than 18 adults, we first find the Z-score for 18.5 (using 18.5 for continuity correction). The Z-score is (X - μ) / σ, where X is 18.5. Thus, Z = (18.5 - 24.75) / 4.27 ≈ -1.46. Looking up this Z-score in the standard normal distribution table or using a calculator, we find the probability of a Z-score being less than -1.46. However, we want the probability of more than 18 adults, so we look for the complement of this, which is 1 - P(Z < -1.46). This gives us a probability of approximately 0.9370, indicating that there is a 93.70% chance that more than 18 adults from the sample would very likely plan to keep a foot in the working world after retirement.

A Survey Conducted in 2011 Asked Employed Adults from Age