Nickie_S
Statistics
The diameters of grapefruits in a certain orchard are normally distributed with a mean of 7.30 inches and a standard deviation of 0.80 inches. Show all work.
(A) What percentage of the grapefruits in this orchard have diameters less than 6.7 inches?
(B) What percentage of the grapefruits in this orchard are larger than 7.20 inches?
What I have wrote out
mean=m=7.30
Standard deviation=s=0.80
(A) what % have diameters less than 6.7
(B) what %are larger than 7.20
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amistre64
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A is asking: P(X<6.7)
amistre64
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B is asking P(X>7.2)
amistre64
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z = (6.7 - mean)/sd
Nickie_S
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6.7-7.30/0.80
amistre64
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good, thats our z score for A
amistre64
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we get a negative and we know that it wants the value of everything to the left of it
amistre64
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.5 - zvalue
amistre64
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zvalue will probably be called a Pvalue later in the book
Nickie_S
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-0.75
amistre64
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yep
mathTalk
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Can we use the bell curve ?
amistre64
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we are .
Nickie_S
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yes, mathtalk that is what they are all made of (im in statistics-the book says elementary statistics- my first class of it0
mathTalk
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somebody show how to proceed here
mathTalk
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please
Nickie_S
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so 0.5000-0.2734(from the table)=
amistre64
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its alot to wrap up form previous posts ... but in a nutshell, you get a zscore and look that up in a table to get a value
amistre64
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then work it into the question to find an answer
amistre64
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.2734 is good, so that should do it for A
amistre64
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B is trickier in that it wants values all to the right of z
amistre64
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same concept, but we have to adjust our thinking a little to accomodate for it
Nickie_S
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don't we do 0.5000-0.2734?
amistre64
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yes
amistre64
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A is the result of .5 - .2734
Nickie_S
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for the first one? right.
amistre64
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yep
Nickie_S
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oh ok, then i have 0.2266
amistre64
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good
Nickie_S
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so then I write it like this right?
P(x<6.7)=0.23
amistre64
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B needs some care taken to it.|dw:1320180861380:dw|
amistre64
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A asks for a % so we gotta rewrite .2266 as 22.66%
Nickie_S
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*100
.2266*100=22.66%
amistre64
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that looks good to me
Nickie_S
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ok, on to the second one!
Nickie_S
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are you still here?
amistre64
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im here
amistre64
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lets figure the z score to get started
Nickie_S
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ok, I have been having loading problems, so I thought it was me.
Nickie_S
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ok
Nickie_S
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P(x>7.2)
Nickie_S
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wait I am trying to figure it.
amistre64
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z = (7.2-7.3)/.8 .... same process, different numbers
Nickie_S
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x-mean/stan dev.
7.20-7.30/0.80
Nickie_S
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-01/0.80
-0.125/0.80
-0.15625
amistre64
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got some typos in there... but your on the right track
-.1/.8 = -.125
amistre64
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7.2 - 7.3 = -.1
-.1/.8 = -1/8 = -.125
Nickie_S
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and then:
-0.125/0.80
=-0.15625
amistre64
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no, there is no need to divide it again by .8
Nickie_S
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ok, i see it. ok next step.
amistre64
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\[z=\frac{x-\bar x}{\sigma}\]
\[z=\frac{7.20-7.30}{0.80}\]
\[z=\frac{-0.10}{0.80}=-0.125\]
amistre64
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id go with the z=.13 value, but a better result would be that since this is exactly in between .12 and .13 to take the average of the values as well
Nickie_S
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lets go with -0.13
amistre64
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lets :)
Nickie_S
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now I go find it in the table.
amistre64
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yep, we need the value for this z to calculate with
Nickie_S
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0.0517
amistre64
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thats what i got in my table as well
Nickie_S
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so do we take 0.5000+0.0517?
amistre64
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we do, and you understand why right?
Nickie_S
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because it is greater, and because it is a positive number
amistre64
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close enough, with more practice it should become better to see it :)
Nickie_S
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lol, I'm a begginer.
amistre64
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one way to think of it is to get the value to the left and subtract it from the total value under the curve; the total is 1
amistre64
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1 - (value to the left of z) = (value to the right of z)
Nickie_S
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because the problem is a greater than, so you have to add the left side of the bell with the amount that is greater than the half.
amistre64
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yep, that is a very good way to view it. good job ;)
Nickie_S
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well thanks for the medal! (medals).
amistre64
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and remember its asking for %
Nickie_S
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so we get 0.5517
Nickie_S
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which makes it a 55.17%
amistre64
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good
Nickie_S
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Ok, I have 2 more questions I need help with, and as I am learning maybe i can do most of the work. So here we go to another question, if your p for it.
amistre64
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im good for it .... :)