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# The table below gives the average monthly temperatures in degrees Fahrenheit for a certain area.Find the mean and median for the following.a. The maximum temperatureb. The minimum temperature

## a. mean $=55.5^{\circ} F \quad$ median $=50.5^{\circ} F$b. mean $=28.92^{\circ} \mathrm{F} \quad$ median $=28.5^{\circ} \mathrm{F}$

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Multivariable Optimization

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this question gives you a list of monthly high and low temperatures and ask you to find the average the mean and the median for both the max end minimum temperatures. So let's start with the max temperature. We knew that the mean is just the sum of all of their data points divided by the number. Clearly, this is over 12 months. A years. We have 12 data points and wouldn't be some. All of our data points up. We get 666 dividing six under 66 By 12 we get 55 0.5. That's her mean temperature over the year. How will we find our median? Will we know when we have 12? Data points are median is the average of X six and X seven. That's the sixth highest and seventh highest data points here. X six is 50 degrees and x seven is 51 degrees, so the average of these to the median is going to be 50 0.5. Now let's do the same thing with the minimum. We know again we have 12 data points, and if we add up all of the minimum temperatures, we get 347 after under 47. Divided by 12 are mean is gonna be 200 in a 28.9. Excuse me, 28.9 is I mean, and now again, we'll find x six and x seven. Those are 26 31 respectively. And so our median is going to be the average of 26 31 which is 28.5 and those your final

University of Oklahoma

#### Topics

Multivariable Optimization

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