anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1363376606947:dw|
anonymous
  • anonymous
but i cant get the right answer
anonymous
  • anonymous
i used that that formula

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More answers

anonymous
  • anonymous
|dw:1363376747863:dw|
anonymous
  • anonymous
z1 = (124-160)/sigma z2 = (196-160)/sigma 1-z1-z2 = (99.7)/100
anonymous
  • anonymous
i have to leave, but I think that should help you find the answer
amistre64
  • amistre64
dont we just need one of those to determine sigma?
amistre64
  • amistre64
\[z=\frac{x-\bar x}{sd}\] \[sd=\frac{x-\bar x}{z}\]
anonymous
  • anonymous
i got confused
anonymous
  • anonymous
why we need to 1- them
amistre64
  • amistre64
or rather, we dont have the z value explicitly do we
anonymous
  • anonymous
we do have z
anonymous
  • anonymous
99.7%
amistre64
  • amistre64
|dw:1363377005023:dw|
amistre64
  • amistre64
we have an implicit z, we can either use the zscore of .3/2 .. or 99.7+.3/2
anonymous
  • anonymous
what would be answer?
amistre64
  • amistre64
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
anonymous
  • anonymous
yes
amistre64
  • amistre64
id prolly use the 196 version
amistre64
  • amistre64
can you tell me the zscore of .9985?
anonymous
  • anonymous
2.96
amistre64
  • amistre64
i agree soo, sd= (196-160)/2.96
anonymous
  • anonymous
oh, i got it
anonymous
  • anonymous
can we use both of them?
amistre64
  • amistre64
both of what?
anonymous
  • anonymous
z1 = (124-160)/sigma z2 = (196-160)/sigma
anonymous
  • anonymous
like that
amistre64
  • amistre64
we can, the sd will be negative for the z1 equation and the percentage at z1 is (.0015), iinstead of (.9985)
anonymous
  • anonymous
thats wht i used initially
anonymous
  • anonymous
but i got wrong answer
amistre64
  • amistre64
one chk would be to do (124-160)/12.1622 and see if its about -.0015 :)
amistre64
  • amistre64
hmm, i get -2.96 is 160 by chance the average of 124 and 196?
amistre64
  • amistre64
.... yep

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