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You are conducting a study of the relationship between the amount of rain in a field and the total mass of fruit produced by tomato plants. You randomly select tomato plants from a field and weigh all the tomatoes on each plant. You plot the data as a histogram, and the data appear to be normally distributed. You calculate an average weight of 2.6 kg of tomatoes per plant, with a variance of 0.5 kg2. a) You have sampled 50 plants. If you assume your data are normally distributed, what is the probability that the 51st plant you sample has 2.6 kg of tomatoes?

b) What is the probability that the weight of tomatoes on the 51st tomato plant is within +/- one standard deviation of the mean weight of your sample? c) Draw a chart that illustrates the probability the 51st tomato plant has greater than 1.2 kg but less than 3.4 kg total weight of tomatoes? The shape of the function can be approximate.

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a) Your distribution here is continuous so the probability of getting a specific exact value is zero. There are two ways of getting intuition behind why this is true. First, there are infinitely many possible values that can come out next, so no matter how common a value is there are too many other values that overwhelm the sample space. The second, if you have seen integration techniques for calculating probabilities notice that they are asking for $\int_{2.6}^{2.6} f(x)$ which is zero since you are integrating over a point (i.e. taking the area under the normal distribution curve over 1 point)

b) This is a fact about normal distributions that 0.6826 is the probability of landing within 1 standard deviation of the norm. I refer you to the z-distribution table, where you can plug in for $z=1.0$ and subtract what you get for $z=-1.0$ http://lilt.ilstu.edu/dasacke/eco148/ztable.htm

c) If Variance is 0.5, standard deviation is 0.707 so they are asking you the probabilty of landing within $(1.2/0.707)$ standard deviations below and $(0.9/0.707)$ standard deviations above. Plug the latter value into the z-table and subtract what you get when you plug the former value into the z-table.

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I wonder if for part a) you can calculate the probability using a probability density function? –  user63036 Feb 21 '13 at 3:45
    
Additionally, I wonder why in part c) you used (0.9/0.707) instead of (3.4/0.707). I'm sorry, I got confuse! –  user63036 Feb 21 '13 at 4:35
    
You can! However since the distribution is continuous you need to integrate over the interval that you want to find the probability of landing in, but since your "interval" is just a point the integral will come out to be zero. –  Sean Ballentine Feb 21 '13 at 4:35
    
I was unclear, its my fault. First, I made a mistake and it should be 0.8 and 0.8 is the distance from the mean (3.4-2.6) so 0.8/0.707 is how many standard deviations from the mean you are. That is what the z-table gives you. The z table gives you "The probability of being less than z standard deviations from the mean". –  Sean Ballentine Feb 21 '13 at 4:38
    
Thanks a lot it makes more sense now!!! –  user63036 Feb 21 '13 at 5:17
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