In: Statistics and Probability
You were told that the mean score on a statistics exam is 75 with the scores normally distributed. In addition, you know the probability of a score between 55 and 60 is 4.41% and that the probability of a score greater than 90 is 6.68%. What is the probability of a score between 55 and 95?
From the given information we have –
Pr (x > 90) = 0.0668,
Pr (z > z1 ) = 0.0668
Pr ( z < z1) = 1 - Pr (z > z1 )
= 1.0 - 0.0668 = 0.9332
From the standard normal table , the 0.9332 Probability is at z = 1.50.
Therefore z = (x – mean) / sd
1.5 = (90 – 75) / sd
1.5 = 15 / sd
Sd = 10
μ=75, σ=10
We need to compute Pr(55≤X≤95). The corresponding z-values needed to be computed are:
Therefore, we get:
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