# The normal distribution I and II Responses

Provide (2) 150 words response for RESPONSES 1 AND 2 below. Responses may include direct questions. In your peer posts, compare the probabilities that you found with those of your classmates. Were they higher/lower and why? In your responses, refer to the specific data from your classmates’ posts. Make sure you include your data set in your initial post as well. Attached are the excel docs for both responses to help with the post.
RESPONSE 1:
This week we worked with averages, standard deviations, and especialy probabilities.
The first step was to calculate a new standard deviation for a sample size of 4. In my case 14264/SQRT(4).
With this number and the previously calculated mean price of a vehicle, in my example 28232, we first calculate the percent chance that the next four vehicles will be 500 dollars below the mean. I came to a 47.2% chance.
The next probability we calculate is the odds of the next four cars being 1000 dollars above the mean. I came to a 44.4% chance this would be the case.
After that we calculate the odds that the next four cars would cost the same as my mean price. I came to a 50% chance that that would happen.
The final probability to calculate was if the next four cars would cost within 1500 dollars +- of the mean price. I came to a 16.7% chance that would happen.
Speaking for myself I found this post quite challenging and would welcome any critical eyes on work.
RESPONSE 2:
For this week’s forum we are asked to find the normal distribution of a set of vehicles and different probabilities.
The mean of all my vehicles without the supercar is 18,478 and I have a new standard deviation of 685.3123781.
The first question asks for the probability that the price will be less than \$500 dollars below the mean. To figure this out we take my mean and subtract 500 dollars to get 17978. That is p(x<17978). In excel make sure to use norm.dist with a formula of TRUE.  Secondly we are asked to find the probability that the price will be higher than \$1000 dollars above the mean. To figure this out we take the mean and add 1000 dollars to get 19478. That is p(x19478). In excel make sure to use norm.dist with a formula of TRUE.  Next we are asked to find the probability that the price will be equal to the mean. To figure this out we take the mean and equal it out against itself at 18478. That is p(x=18478). In excel make sure to use norm.dist with a formula of FALSE.  Finally we are asked to find the probability that the price will be \$1500 within the mean. To figure this out we take the mean 18478, and subtract 1500 as well as add 1500 in a separate equation. Within excel I input the normal distribution formula with the inputed numbers of =NORM.DIST(19978,16978,C28,TRUE)-NORM.DIST(16978,18478,C28,TRUE).

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