Roger's method would not provide relevant results for what he was trying to accomplish because he was calculating average over a 3 month period instead of for a single day.
<h3>How to find the mean temperature?</h3>
Mean/Average of any given data simply means the sum of the given data divided by the sample size of the given data.
Now, when talking about average temperature, it simply means adding the temperatures for the duration you intend to analyze and then dividing the sum by the number of temperatures you added.
However, in this question Roger wants to find the average temperature for any given data and we know the temperatures in a 24-hour day may differ from time to time and as such the mean/average for a particular day would simply be found by adding up temperatures for the day and then dividing by the number of temperatures added.
Thus, Roger's method would not provide relevant results for what he was trying to accomplish because he was calculating average over a 3 month period instead of for a single day.
Read more about Mean Temperature at; brainly.com/question/1456846
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Answer:
P= 3.27
Step-by-step explanation:
Hope this helps sorry for the crumbled paper
Answer:
A. 2[24÷ (4+4)]
Step-by-step explanation:
Hope this helps :)
Let Xi be the random variable representing the number of units the first worker produces in day i.
Define X = X1 + X2 + X3 + X4 + X5 as the random variable representing the number of units the
first worker produces during the entire week. It is easy to prove that X is normally distributed with mean µx = 5·75 = 375 and standard deviation σx = 20√5.
Similarly, define random variables Y1, Y2,...,Y5 representing the number of units produces by
the second worker during each of the five days and define Y = Y1 + Y2 + Y3 + Y4 + Y5. Again, Y is normally distributed with mean µy = 5·65 = 325 and standard deviation σy = 25√5. Of course, we assume that X and Y are independent. The problem asks for P(X > Y ) or in other words for P(X −Y > 0). It is a quite surprising fact that the random variable U = X−Y , the difference between X and Y , is also normally distributed with mean µU = µx−µy = 375−325 = 50 and standard deviation σU, where σ2 U = σ2 x+σ2 y = 400·5+625·5 = 1025·5 = 5125. It follows that σU = √5125. A reference to the above fact can be found online at http://mathworld.wolfram.com/NormalDifferenceDistribution.html.
Now everything reduces to finding P(U > 0) P(U > 0) = P(U −50 √5125 > − 50 √5125)≈ P(Z > −0.69843) ≈ 0.757546 .
