Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 401 drivers and find that 294 claim to always buckle up. Construct a 90% confidence interval for the population proportion that claim to always buckle up. Use interval notation, for example, [1,5].

[0.6969, 0.7695]

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

They randomly survey 401 drivers and find that 294 claim to always buckle up.

This means that , z is the value of Z that has a p-value of , so .  

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for the population proportion that claim to always buckle up is [0.6969, 0.7695]

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.