## anonymous 3 years ago 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?

1. anonymous

|dw:1363376606947:dw|

2. anonymous

but i cant get the right answer

3. anonymous

i used that that formula

4. anonymous

|dw:1363376747863:dw|

5. anonymous

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

6. anonymous

7. amistre64

dont we just need one of those to determine sigma?

8. amistre64

$z=\frac{x-\bar x}{sd}$ $sd=\frac{x-\bar x}{z}$

9. anonymous

i got confused

10. anonymous

why we need to 1- them

11. amistre64

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

12. anonymous

we do have z

13. anonymous

99.7%

14. amistre64

|dw:1363377005023:dw|

15. amistre64

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

16. anonymous

17. 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

18. anonymous

yes

19. amistre64

id prolly use the 196 version

20. amistre64

can you tell me the zscore of .9985?

21. anonymous

2.96

22. amistre64

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

23. anonymous

oh, i got it

24. anonymous

can we use both of them?

25. amistre64

both of what?

26. anonymous

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

27. anonymous

like that

28. amistre64

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

29. anonymous

thats wht i used initially

30. anonymous

31. amistre64

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

32. amistre64

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

33. amistre64

.... yep