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Weights of apples grown in an orchard are known to follow a normal distribution with mean 160 grams. It is known that approximately 99.7% of apples have weights between124 and 196 grams. What is the standard deviation of weights of all apples grown in the orchard?

Mathematics
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|dw:1363376606947:dw|
but i cant get the right answer
i used that that formula

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Other answers:

|dw:1363376747863:dw|
z1 = (124-160)/sigma z2 = (196-160)/sigma 1-z1-z2 = (99.7)/100
i have to leave, but I think that should help you find the answer
dont we just need one of those to determine sigma?
\[z=\frac{x-\bar x}{sd}\] \[sd=\frac{x-\bar x}{z}\]
i got confused
why we need to 1- them
or rather, we dont have the z value explicitly do we
we do have z
99.7%
|dw:1363377005023:dw|
we have an implicit z, we can either use the zscore of .3/2 .. or 99.7+.3/2
what would be answer?
does that make sense? 99.7% is inbetween those values, so the proper zscore has to be adjusted. use z(.9985), at x=196 or, z(.0015), at x=124
yes
id prolly use the 196 version
can you tell me the zscore of .9985?
2.96
i agree soo, sd= (196-160)/2.96
oh, i got it
can we use both of them?
both of what?
z1 = (124-160)/sigma z2 = (196-160)/sigma
like that
we can, the sd will be negative for the z1 equation and the percentage at z1 is (.0015), iinstead of (.9985)
thats wht i used initially
but i got wrong answer
one chk would be to do (124-160)/12.1622 and see if its about -.0015 :)
hmm, i get -2.96 is 160 by chance the average of 124 and 196?
.... yep

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