Can someone help me with both these problems?? Thx!
1 answer:
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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:
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
We have to evaluate the probability that more than 47 cars carry just one person.
After continuity correction, we will evaluate
Calculation the value from standard normal z table, we have,
0.996 is the probability that more than half out of 92 vehicles carry just one person.