Vincent M. answered • 11/10/24
Patient and Collaborative Teacher and Learner-- Vincent
Hello Chad,
Let me run through each part step-by-step to help you understand what is happening here.
(A) Part A is only testing your knowledge about Uniform Distribution, we know this because the central limit theorem will only come into play when our sample size "n" is greater than 30, and in this case, it is only 1.
To solve for the Probability P such that (P>-5.2) we can take the proportion of temperature in the range of values that are greater than -5.2 over all of the values over the range. This equation looks like the following:
((-4)- (-5.2))/(-4)-(-6) = 1.2/2 = .6. Therefore (P>-5.2) =.6.
We were able to use this proportion because the data is uniformly distributed meaning that when only this one sample was taken, all values in the range are equally likely.
Part B and C will both utilize the Central Limit Theorem (CLT) since they have samples greater than the 30-sample threshold at 50 and 40 respectively. Any distribution that can be described by the central limit theorem means that it is being described by a normal distribution. To begin each part we will find the mean and standard deviation of this normal distribution.
(B) We will first need to solve for the mean which can be gathered by the equation (A+B)/2 where a and B are the original rate limits (-4 and -6 in this case). This will give us a mean =-5
Next, we can solve for the standard deviation which uses the equation (A+B)^2/12. Using the same A and B from above we get that the standard deviation is (1/3).
We must remember though that this does not fully describe the normal distribution of this function, instead, we need to account for the 50 samples that were taken. According to the CLT, this will not affect the mean but only affect the standard deviation. To get the standard deviation we will take sqrt(1/3)/sqrt(50) or in more general terms sqrt(original standard deviation)/sqrt(number of samples "n"). We then get this answer to be about .0913 in our case.
Next, we can use a normal model to predict the probabilities. In this case, we are seeing the probability the average is above -5.2. We can set up a z-score calculation for our normal model with z = -5.2 - (-5)/ .0913, returning a z= -2.19. The number -5.2 comes from the number we are comparing from, -5 is the average that we calculated above, and .0913 is the standard deviation of the size 50 sample.
Using the z score we can either use a normal distribution table (can be found online) or follow along with my TI-84 Calaculator Tutorial to find the Probability):
On Ti-84: 1) With calculator on press the 2nd key, then press the vars key (located under arrows), 2) navigate down to option number 2: normalcdf(, press enter into this option. 3) set lower: to our z score of -2.19, arrow down to upper and set to a large number (100 will suffice), next set the u looking symbol (mu) to 0 and the curisve o looking symbol (sigma) to 1, finally press enter on paste, the following decimal value of .9857 represent the probability we are looking for.
(C) Part C is very similar to part B so we will run through it quicker mentioning only different steps.
The mean will still be -5
Since the sample is smaller at only 40 the samples standard deviation will be .1041
We can solve for both the z score of -4.9 and -5.1 which are respecitlvley .96 and -.96
You can do one of two things, we can either use a normal distribution table to subtract the probability that z=.96 - probability that z=-.96, or we can use the TI-84 calculator function setting the lower limit to -.96 and upper to .96, both instances will return the correct probability of .663.
I hope this helped and if you had any issues following mu logic or need further explanations please let me know.
Mr. Vincent!
