Answer:
The probability that at least two-third of vehicles in the sample turn is 0.4207.
Explanation:
Let X = number of vehicles that turn left or right.
The proportion of the vehicles that turn is, p = 2/3.
The nest n = 50 vehicles entering this intersection from the east, is observed.
Any vehicle taking a turn is independent of others.
The random variable X follows a Binomial distribution with parameters n = 50 and p = 2/3.
But the sample selected is too large and the probability of success is close to 0.50.
So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:
- np ≥ 10
- n(1 - p) ≥ 10
Check the conditions as follows:
![np=50* (2)/(3)=33.333>10\\\\n(1-p)=50* (1)/(3)= = 16.667>10](https://img.qammunity.org/2021/formulas/mathematics/college/8jbnyk8cfjefyi6ad45nr9v1g1mgqduqar.png)
Thus, a Normal approximation to binomial can be applied.
So,
![X\sim N(np, np(1-p))](https://img.qammunity.org/2021/formulas/mathematics/college/vv8jkj82iikquqkp2bh5o63yux15q8hrmz.png)
Compute the probability that at least two-third of vehicles in the sample turn as follows:
![P(X\geq (2)/(3)* 50)=P(X\geq 33.333)=P(X\geq 34)](https://img.qammunity.org/2021/formulas/mathematics/college/wu8v1m074rhg2vn4so7zpvmya8tmqxq7b9.png)
![=P((X-\mu)/(\sigma)>\frac{34-33.333}{\sqrt{50* (2)/(3)*\frac {1}{3}}})](https://img.qammunity.org/2021/formulas/mathematics/college/3x00ijl2lpmdaflgatcsls8kbs7y6sjgja.png)
![=P(Z>0.20)\\=1-P(Z<0.20)\\=1-0.5793\\=0.4207](https://img.qammunity.org/2021/formulas/mathematics/college/3l8gpjj9cbw4t2ayulmdvrawavddsfaff8.png)
Thus, the probability that at least two-third of vehicles in the sample turn is 0.4207.