Answer:
a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
b) 0.996 is the probability that more than half of the vehicles carry just one person.
Step-by-step explanation:
We are given the following information:
A) Binomial distribution
We treat vehicle on road with one passenger as a success.
P(success) = 64% = 0.64
Then the number of vehicles follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 10
We have to evaluate:
![P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291](https://tex.z-dn.net/?f=P%28x%20%5Cgeq%206%29%20%3D%20P%28x%20%3D6%29%20%2B...%2B%20P%28x%20%3D%2010%29%20%5C%5C%3D%20%5Cbinom%7B10%7D%7B6%7D%280.64%29%5E6%281-0.64%29%5E4%20%2B...%2B%20%5Cbinom%7B10%7D%7B10%7D%280.64%29%5E%7B10%7D%281-0.79%29%5E0%5C%5C%3D0.7291)
0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.
B) By normal approximation
Sample size, n = 92
p = 0.64
![\mu = np = 92(0.64) = 58.88](https://tex.z-dn.net/?f=%5Cmu%20%3D%20np%20%3D%2092%280.64%29%20%3D%2058.88)
![\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B92%280.64%29%281-0.64%29%7D%20%3D%204.60)
We have to evaluate the probability that more than 47 cars carry just one person.
![P(x \geq 47)](https://tex.z-dn.net/?f=P%28x%20%5Cgeq%2047%29)
After continuity correction, we will evaluate
Calculation the value from standard normal z table, we have,
![P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%](https://tex.z-dn.net/?f=P%28x%20%3E%2046.5%29%20%3D%201%20-%200.004%20%3D%200.996%20%3D%2099.6%5C%25)
0.996 is the probability that more than half out of 92 vehicles carry just one person.