Answer:
The probability that at least two-third of vehicles in the sample turn is 0.4207.
Step-by-step explanation:
Let <em>X</em> = number of vehicles that turn left or right.
The proportion of the vehicles that turn is, <em>p</em> = 2/3.
The nest <em>n</em> = 50 vehicles entering this intersection from the east, is observed.
Any vehicle taking a turn is independent of others.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 50 and <em>p</em> = 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 <em>X</em> if the following conditions are satisfied:
- np ≥ 10
- n(1 - p) ≥ 10
Check the conditions as follows:
![np=50\times \frac{2}{3}=33.333>10\\\\n(1-p)=50\times \frac{1}{3}= = 16.667>10](https://tex.z-dn.net/?f=np%3D50%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%3D33.333%3E10%5C%5C%5C%5Cn%281-p%29%3D50%5Ctimes%20%5Cfrac%7B1%7D%7B3%7D%3D%20%3D%2016.667%3E10)
Thus, a Normal approximation to binomial can be applied.
So, ![X\sim N(np, np(1-p))](https://tex.z-dn.net/?f=X%5Csim%20N%28np%2C%20np%281-p%29%29)
Compute the probability that at least two-third of vehicles in the sample turn as follows:
![P(X\geq \frac{2}{3}\times 50)=P(X\geq 33.333)=P(X\geq 34)](https://tex.z-dn.net/?f=P%28X%5Cgeq%20%5Cfrac%7B2%7D%7B3%7D%5Ctimes%2050%29%3DP%28X%5Cgeq%2033.333%29%3DP%28X%5Cgeq%2034%29)
![=P(\frac{X-\mu}{\sigma}>\frac{34-33.333}{\sqrt{50\times \frac{2}{3}\times\frac {1}{3}}})](https://tex.z-dn.net/?f=%3DP%28%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cfrac%7B34-33.333%7D%7B%5Csqrt%7B50%5Ctimes%20%5Cfrac%7B2%7D%7B3%7D%5Ctimes%5Cfrac%20%7B1%7D%7B3%7D%7D%7D%29)
![=P(Z>0.20)\\=1-P(Z](https://tex.z-dn.net/?f=%3DP%28Z%3E0.20%29%5C%5C%3D1-P%28Z%3C0.20%29%5C%5C%3D1-0.5793%5C%5C%3D0.4207)
Thus, the probability that at least two-third of vehicles in the sample turn is 0.4207.