In: Statistics and Probability
Japanese automobiles comprise approximately 20% of all new car sales in the United States. Suppose a state licensing office received 100 requests for license plates for new cars during a given week.
a) What is the probability that 25 or more of the license requests are for Japanese cars?
b) What is the probability that no more than 10 of the requests are for Japanese cars?
c) What is the probability that at least 10, but no more than 20, Japanese cars are represented within the group?
Japanese automobiles comprise approximately 20% of all new car sales in the US.
A state licensing office received 100 requests for license plates for new cars during a given week.
So, if X be the random variable denoting the number of cars out of these 100, which are Japanese cars, then X follows binomial with parameters n=100, and p=0.2.
Now, we note that n is too large to apply binomial dostribution; n*p=100*0.2=20, which is also large for poisson distribution.
So, we have to apply normal approximation to binomial distribution in this problem.
So,
Mean of X
=100*0.2
=20
Standard deviation of X
=Square root of 100*0.2*(1-0.2)
=Square root of 16
=4
So, we can say that X approximately normal distribution with mean 20 and standard deviation of 4.
So, we can standardise this by saying
Z=(X-20)/4 follows standard normal with mean 0 and standard deviation of 1.
Question a
We have to find the probability that 25 or more license requests are for Japanese cars.
Now, this means we have to find
Where, phi is the distribution function of the standard normal variate.
From the standard normal table, this becomes
So, the answer is 0.106.
Question (b)
We have to find the probability that no more than 10 of the cars are Japanese.
So, we have to find
Where, phi is the distribution function of the standard normal variate.
From the standard normal table, this becomes
So, the answer is 0.006.
Question (c)
We have to find the probability that at least 10, but not more than 20 cars, are represented by Japanese automobiles.
So, we have to find
Where, phi is the distribution function of the standard normal variate.
From the standard normal table, it becomes
So, the answer is 0.494.