X= the mean, s= standard deveation
x=30, s=2
what percent of the scores would be found between 28 and 32?
what percent of the scores would be found between 26 and 34?
Mean, = 30
Standard deviation, = 2
The percent of scores that would be found between 28 and 32
= P(28 ≤ X ≤ 32) * 100%
= P{(28 - 30)/2 ≤ Z ≤ (32 - 30)/2} * 100%
=P(-1 ≤ Z ≤ 1)*100%
= 68.26%
The percent of scores that would be found between 26 and 34
= P(26 ≤ X ≤ 34) * 100%
= P(-2 ≤ Z ≤ 2)*100%
= 95.44%
X= the mean, s= standard deveation x=30, s=2 what percent of the scores would be found...
If a score are normally with the mean of 30 and deviation of s, what perce a greater than 30? l'equal to s a greater than 37 dl between do and 34 e normally distribution in of 30 and the standard ors, what percent of the scores is
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