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.
is the probability distribution of x approximately normal? justify your answer.
np = _ and n(1 - p) = _
therefore, the probability distribution of x ___ approximately normal.

Explanation:

Step1: Identify n and p

We have a sample size \( n = 200 \) and the probability of success (using public transport) \( p=0.345 \).

Step2: Calculate \( np \)

To check the normal approximation for a binomial distribution, we first calculate \( np \).
\( np=200\times0.345 = 69\)

Step3: Calculate \( n(1 - p) \)

Next, calculate \( n(1 - p) \). First, find \( 1-p=1 - 0.345=0.655 \). Then \( n(1 - p)=200\times0.655 = 131\)

Step4: Check normal approximation conditions

For a binomial distribution \( X\sim B(n,p) \), it can be approximated by a normal distribution \( N(np,np(1 - p)) \) if \( np\geq10 \) and \( n(1 - p)\geq10 \). Here, \( np = 69\geq10 \) and \( n(1 - p)=131\geq10 \), so the conditions are satisfied.

Answer:

\( np=\boldsymbol{69} \), \( n(1 - p)=\boldsymbol{131} \), and since \( np\geq10 \) and \( n(1 - p)\geq10 \), the probability distribution of \( X \) is approximately normal.