Question

in a large city, 34% of residents use public transportation at least once per week. suppose the mayor selects a random sample of 200 residents. let x = the number of residents in the sample who use public transportation at least once per week. use an appropriate normal distribution to calculate the probability that 60 or fewer residents in the sample use public transportation at least once per week. probability = \boxed{} (round to 4 decimal places. leave your answer in decimal form.)

Explanation:

Step1: Identify distribution parameters

We have a binomial distribution with \( n = 200 \) (sample size) and \( p = 0.34 \) (probability of success, i.e., using public transport). For normal approximation, we check \( np \geq 5 \) and \( n(1 - p) \geq 5 \).
\( np = 200\times0.34 = 68 \), \( n(1 - p)=200\times0.66 = 132 \), both are \( \geq 5 \), so normal approximation is valid.

Step2: Calculate mean and standard deviation

Mean \( \mu = np = 200\times0.34 = 68 \).
Standard deviation \( \sigma=\sqrt{np(1 - p)}=\sqrt{200\times0.34\times0.66}=\sqrt{44.88}\approx6.6993 \).

Step3: Apply continuity correction

We want \( P(X \leq 60) \) for binomial, so with continuity correction, we use \( X \leq 60.5 \) for normal.

Step4: Calculate z - score

\( z=\frac{x-\mu}{\sigma}=\frac{60.5 - 68}{6.6993}=\frac{-7.5}{6.6993}\approx - 1.12 \).

Step5: Find probability from z - table

\( P(Z \leq - 1.12) \). From standard normal table, \( P(Z \leq - 1.12)=0.1314 \).

Answer:

\( 0.1314 \)