according to a survey, 10% of americans are afraid to fly. suppose 1,100 americans are sampled.\na. what is…

Explanation:

Step1: Identify distribution parameters

This is a binomial - to - normal approximation problem. Let (n = 1100) (sample size) and (p=0.1). The mean of the binomial distribution is (\mu = np=1100\times0.1 = 110), and the standard deviation is (\sigma=\sqrt{np(1 - p)}=\sqrt{1100\times0.1\times(1 - 0.1)}=\sqrt{1100\times0.1\times0.9}=\sqrt{99}\approx9.95).

Step2: Standardize for part a

For (x = 121), the z - score is (z=\frac{x-\mu}{\sigma}=\frac{121 - 110}{9.95}=\frac{11}{9.95}\approx1.11). We want (P(X\geq121)), which is equivalent to (P(Z\geq1.11)) in the normal - approximation. Using the standard normal table, (P(Z\geq1.11)=1 - P(Z < 1.11)=1 - 0.8665 = 0.1335), or (13.35%).

Step3: Standardize for part b

For (x = 165), the z - score is (z=\frac{x-\mu}{\sigma}=\frac{165 - 110}{9.95}=\frac{55}{9.95}\approx5.53). We want (P(X\geq165)), which is equivalent to (P(Z\geq5.53)) in the normal - approximation. Since the standard normal table typically only goes up to (z = 3.49), and for (z\geq3.49), (P(Z\geq z)\approx0). So (P(Z\geq5.53)\approx0.00%).

Step4: Standardize for part c

The number of Americans in the sample corresponding to (8%) is (x = 0.08\times1100 = 88). The z - score is (z=\frac{x-\mu}{\sigma}=\frac{88 - 110}{9.95}=\frac{- 22}{9.95}\approx - 2.21). We want (P(X\leq88)), which is equivalent to (P(Z\leq - 2.21)) in the normal - approximation. Using the standard normal table, (P(Z\leq - 2.21)=0.0136), or (1.36%).

Answer:

a. (13.35%) b. (0.00%) c. (1.36%)