Rahman W.

asked • 07/27/20

Data Management Normal Distribution Probability

The heights of 16-month-old oak seedlings are normally distributed with a mean of 31.5 cm and a standard deviation of 10 cm. What is the range of heights between which 5 % of the seedlings will grow?

i did

μ = 31.5 cm 

o = 10cm

5% = 95% - 90% 

Z - Score *Found using the z - score table*: 

95% = 1.64 and 90% = -1.64 

31.5 + 1.64(10) and 31.5 - 1.64(10) 

47.9 and 15.1

but someone else got 25.76 and 26.24. How do I get to that answer? Who's right?

Mike D.

Rahman. You started well , but your second Z score is wrong, it should be 1.28 approximately. That gives a range for you of 44.3 to 47.9. This is a valid answer. In fact there are many possible answers. As long as the difference in areas is 5% as far as I can see the height range is valid. There isn't a unique range of heights, unless some further constraint is satisfied.


1 Expert Answer


Mawais F. answered • 08/05/20

New to Wyzant

Ungrad Suny Albany 2019

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