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?

- anonymous

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- anonymous

|dw:1363376606947:dw|

- anonymous

but i cant get the right answer

- anonymous

i used that that formula

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## More answers

- anonymous

|dw:1363376747863:dw|

- anonymous

z1 = (124-160)/sigma
z2 = (196-160)/sigma
1-z1-z2 = (99.7)/100

- anonymous

i have to leave, but I think that should help you find the answer

- amistre64

dont we just need one of those to determine sigma?

- amistre64

\[z=\frac{x-\bar x}{sd}\]
\[sd=\frac{x-\bar x}{z}\]

- anonymous

i got confused

- anonymous

why we need to 1- them

- amistre64

or rather, we dont have the z value explicitly do we

- anonymous

we do have z

- anonymous

99.7%

- amistre64

|dw:1363377005023:dw|

- amistre64

we have an implicit z, we can either use the zscore of .3/2 .. or 99.7+.3/2

- anonymous

what would be answer?

- 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

yes

- amistre64

id prolly use the 196 version

- amistre64

can you tell me the zscore of .9985?

- anonymous

2.96

- amistre64

i agree
soo, sd= (196-160)/2.96

- anonymous

oh, i got it

- anonymous

can we use both of them?

- amistre64

both of what?

- anonymous

z1 = (124-160)/sigma
z2 = (196-160)/sigma

- anonymous

like that

- amistre64

we can, the sd will be negative for the z1 equation
and the percentage at z1 is (.0015), iinstead of (.9985)

- anonymous

thats wht i used initially

- anonymous

but i got wrong answer

- amistre64

one chk would be to do (124-160)/12.1622 and see if its about -.0015 :)

- amistre64

hmm, i get -2.96
is 160 by chance the average of 124 and 196?

- amistre64

.... yep

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