The z-probability for the condition -0.78 ≤ z ≤ 1.16 is 66%.
What is a z-score?
A z-score is a numerical measurement that describes a value's relationship to the mean of a group of values.
It is given that
-0.78 ≤ z ≤ 1.16
We have to find the p-value corresponding to -0.78 ≤ z ≤ 1.16.
This means, p(-0.78 ≤ z ≤ 1.16)
p(0.78≤ z ≤1.16) can be written as p(z ≤ 1.16) - p(z ≤ -0.78)
p(0.78≤ z ≤1.16) = p(z ≤ 1.16) - (1-p(z ≤ 0.78))
p(0.78≤ z ≤1.16) = p(z ≤ 1.16) -1 + p(z ≤ 0.78)
From the standard normal table,
p(z ≤ 1.16)=0.87698
p(z ≤ 0.78)=0.78230
So, p(0.78≤ z ≤1.16)= 0.87698-1+0.78230
p(0.78≤ z ≤1.16) = 0.65928≈0.66 i.e 66%
Therefore, the z-probability for the condition -0.78 ≤ z ≤ 1.16 is 66%.
To get more about standard normal distribution visit:
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