Denise S.

asked • 10/02/22

Statistics math

The Weight of a one yr old girls in the United States are approximately normally distributed with a Mean of 21.1 pounds and a standard deviation of 2.8 pounds, determine the proporion of 13.1 pounds or less

Christina P.

In order to solve this, you must first find the z-score for 13.1 pounds. A z-score is equal to the difference between the individual score (which in this case is 13.1) and the mean (which in this case is 21.1) and dividing that value by the standard deviation (which in this case is 2.8). Thus, we get (13.1-21.1)/2.8 which is approximately -2.8571. This is a negative number because the individual is lighter than the mean. Now, we need to find the proportion of the disribution that is below a z-score of -2.8571 using a z-score table. If this is for a class, you will have a table like this provided by your teacher or in your textbook. By checking a z-score table we would see that the area below this z-score (which can also be referred to as "area to the lower tail") is about 0.22%. Keep in mind that it is always best to consider whether your answer makes sense as a way to double-check that you've done your calculations correctly. In this example, we are saying less than 1% of people would be lighter than someone who is 13.1 pounds. This is a very small proportion of people which is what we would expect because we are looking for the proportion of people who are lighter than someone who is already much lighter than the average person. The farther an individual score is from the mean, the rarer it is expected to be.
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01/21/23

Judy R.

Hello I am 22 years experienced Online Tutor and Assignment Helper for Computer Science and Math. I have been teaching IT Professionals, students from different grades, and Graduate and Post Graduate students for more than 22 years. I am ready to help you with your learning requirements, assignments, tests, and projects. For further discussion about the assignment or project help you need, please add me on skype and my skype id is nettuitions Thanks
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01/22/23

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