Daily Commute About 83% of U.S. employees drive their own vehi...

Question

Daily Commute About 83% of U.S. employees drive their own vehicle to work. You randomly select a sample of U.S. employees. Find the probability that more than 100 of the employees drive their own vehicle to work. (Source: U.S. Bureau of Labor Statistics)

c. You select 150 U.S. employees.

c. You select 150 U.S. employees.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A recent survey found that 76% of adults in a city use public transportation at least once a week. Suppose you randomly select a sample of 120 adults from the city. What is the probability that more than 80 of them use public transportation at least once a week? Awesome. So it appears for this particular problem, we're asked to determine what the probability value is that more than 80 of these 120 adults selected for for sampling for a survey use public transportation at least once a week. So now that we know we're ultimately trying to determine what this probability value is. Our first First thing we wanna do is let's read off our multiple choice answers and see what our final answer might be, now that we know that we're ultimately once again trying to determine the probability. So A is 0.0111, B is 0.0333, C is 0.9667, and finally, D is 0.9889. Awesome. So first off, let us define the problem as a binomial probability. And let us state that X will be the number of adults in the sample who use public transportation at least once a week. And let us note that X follows a binomial distribution, meaning that we could say that X, which is a binomial, which I'll call B for short. Which is lowercase n is equal to, so N being our sample size is equal to 120. And P lowercase p is equal to 0. 76 where P is our percentage, and we wrote our percentage, which was given to us as 76% as a decimal. Fantastic. So now we need to state what we are being asked to solve for, so we're trying to find the probability P of X is greater than 80. So with that in mind, as we should recall, we need to check if the normal approximation is appropriate for the case of this problem, and we need to note that since N, as we should recall, is larger and P is not too close to 0 or 1, we can use the normal approximation. So with that in mind, we can, we can now calculate the mean. Which the mean will denote as mu, and mu, the mean is equal to N multiplied by P, which is equal to 120 multiplied by. 0.76, which will give us an answer of 91.2. We also need to calculate the standard deviation, which we should recall that sigma, the standard deviation, is equal to the square root of N multiplied by P, multiplied by parentheses 1 minus P, which is equal to the square root of 120 multiplied by 0.76 multiplied by parenthesis, so multiplied by parentheses. 1 minus our value for P, which is 0.76. So when we plug that into our calculator, we will find that our standard deviation sigma will be equal to the square root of, let me round to three decimal places, 21.888. Fantastic. So, moving right along here, we now need to recall and apply the continuity correction and With that in mind, we need to note that for PX is greater than 80, we need to use PX is greater than or equal to 81, and then we need to apply our continuity correction, which is P X is greater than or equal to. 80.5. Fantastic. So our next step now is we need to convert to a Z score by recalling and using our Z score equation which states that Z, the Z score is equal to for the case of this problem, 80.5 minus 91.2, divided by the square root of 21.888. Which will be approximately equal to -2.287077, and this is when we round to 123456 decimal places. So once again, this is when we round to 6 decimal places. So, now, our next step is we need to find the probability using the standard normal distribution table, and you should have a standard normal distribution table in your textbook. So let me use it. We will find that the probability P of X is greater than 80, is equal to P of Z is greater than -2.2. 87077 and we need to note that Z is equal to -2.287077 is in between -2.28 and 2.29. So when we use our standard normal distribution table, we will be able to say that P of Z is less than -2.28, is approximately equal to 0.0113, and that P of Z is less than -2.29 will be approximately equal to 0.0110. Fantastic, moving right along here. So, when we recall by interpolarization, Or interpolation, we will find that P of Z is less than -2.287077 will be equal to -2.2. 87077 minus -2.28. All divided by -2.29 minus -2.28. And then we need to multiply everything by parentheses 0.0, so 0.0110 minus 0.0113. Plus 0.0113, which when we perform that calculation, we will find that it's approximately equal to 0.0111, when we round to 4 decimal places. So therefore, without a doubt, we could go ahead and write that P of Z is greater than -2.287077 will be equal to 1 minus P of Z is less than. Less than 2.2. 87077, which will mean that this is equal to 1 minus 0.0111, which when we plug that into our calculator, we will find that it will be equal to when we round to 4 decimal places to match all of our multiple choice answers because they're all rounded to 4 decimal places. We will find that it's equal to 0.9. 889, and that is our final answer. All right, we did it. So the probability that more than 80 of the 120 adults use public transportation at least once a week will be approximately 0.9889. And looking at our multiple choice answers, our correct answer without a doubt has to be multiple choice answer letter D. Which once again is 0.9889. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Daily Commute About 83% of U.S. employees drive their own vehicle to work. You randomly select a sample of U.S. employees. Find the probability that more than 100 of the employees drive their own vehicle to work. (Source: U.S. Bureau of Labor Statistics)

c. You select 150 U.S. employees.

Key Concepts

Probability

Binomial Distribution

Normal Approximation

Watch next

Daily Commute About 83% of U.S. employees drive their own vehicle to work. You randomly select a sample of U.S. employees. Find the probability that more than 100 of the employees drive their own vehicle to work. (Source: U.S. Bureau of Labor Statistics)c. You select 150 U.S. employees.

Key Concepts

Probability

Binomial Distribution

Normal Approximation

Watch next

Daily Commute About 83% of U.S. employees drive their own vehicle to work. You randomly select a sample of U.S. employees. Find the probability that more than 100 of the employees drive their own vehicle to work. (Source: U.S. Bureau of Labor Statistics)

c. You select 150 U.S. employees.