The distribution of heights of 5-year old children (in centimeters) is N(100, 62) .
The middle 80% of all heights of 5 year old children fall between ____ and ____. (use 2 decimal places).
I got 92.29 and 107.71. I just wanted to confirm it with someone. Thanks!
The middle 80% of all heights of 5 year old children fall between ____ and ____. (use 2 decimal places).
I got 92.29 and 107.71. I just wanted to confirm it with someone. Thanks!
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Kumar -
The middle 80% will have z-values where the area under the standard normal curve equals 10% in each tail. Agree?
Next, look up the z-values that equate to 10% tails:
z-value = +/-1.282
Finally, set up your interval:
100 +/- (1.282)(sqrt62) = 89.9 to 110.09
Hope that helped
The middle 80% will have z-values where the area under the standard normal curve equals 10% in each tail. Agree?
Next, look up the z-values that equate to 10% tails:
z-value = +/-1.282
Finally, set up your interval:
100 +/- (1.282)(sqrt62) = 89.9 to 110.09
Hope that helped
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I'm assumig the "62" represents the variance and not the std dev.
from normal tables, z-score cutting off 10% in each tail is 1.28
so (X-100)/sqrt(62) = 1.28; solving for X, I get X=110.08
and by symmetry, I would get 89.92 in the left tail.
Check the z-score "cut point" you used.
from normal tables, z-score cutting off 10% in each tail is 1.28
so (X-100)/sqrt(62) = 1.28; solving for X, I get X=110.08
and by symmetry, I would get 89.92 in the left tail.
Check the z-score "cut point" you used.