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. (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 calculate the mean \( \mu = np \) and standard deviation \( \sigma=\sqrt{np(1 - p)} \).

\( \mu = 200\times0.34 = 68 \)

\( \sigma=\sqrt{200\times0.34\times(1 - 0.34)}=\sqrt{200\times0.34\times0.66}=\sqrt{44.88}\approx6.6993 \)

Step2: Apply continuity correction

We want \( P(X\leq60) \). For normal approximation, we use continuity correction, so we find \( P(X\leq60.5) \).

Step3: Calculate z - score

The z - score is calculated as \( z=\frac{x-\mu}{\sigma} \), where \( x = 60.5 \), \( \mu = 68 \), and \( \sigma\approx6.6993 \).

\( z=\frac{60.5 - 68}{6.6993}=\frac{-7.5}{6.6993}\approx - 1.12 \)

Step4: Find the probability from z - table

We need to find \( P(Z\leq - 1.12) \). Looking up in the standard normal table, \( P(Z\leq - 1.12)=0.1314 \)

Answer:

\( 0.1314 \)