Suppose a research firm conducted a survey to determine the average amount of money steady smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed that the sample mean is $20 and the sample standard deviation is $5. What is the probability that a sample of 100 steady smokers spend between $19 and $21? a.0.4772 b.0.0228 c.0.9544 d.$20

Answer:option (b) 0.0228Step-by-step explanation:Data provided in the question:Sample size, n = 100Sample mean, μ = $20Standard deviation, s = $5Confidence interval = between $19 and $21Now,Confidence interval = μ ± [tex]z\frac{s}{\sqrt n}[/tex]thus, Upper limit of the Confidence interval = μ + [tex]z\frac{s}{\sqrt n}[/tex]or$21 =  $20 + [tex]z\frac{5}{\sqrt{100}}[/tex]orz = 2Now, P(z = 2) = 0.02275                       [From standard z vs p value table] orP(z = 2) ≈ 0.0228Hence, the correct answer is option (b) 0.0228