(1 point) Suppose that X is normally distributed with mean 120 and standard deviation 14.
A. What is the probability that X is greater than 140.58? Probability =
B. What value of X does only the top 14% exceed? X =
Solution:
Given that,
= 120
= 14
A) p ( x > 140.58 )
= 1 - P ( x < 140.58 )
= 1 - P ( ( x - / ) < ( 140.58 - 120 / 14 ) )
= 1 - P ( z < 20.58/14 )
= 1 - p ( z < 1.47 )
Using z table
= 1 - 0.9292
= 0.0708
Probability = 0.0708
B ) P(Z > z) = 14%
1 - P(Z < z) = 0.14
P(Z < z) = 1 - 0.14 = 0.86
P(Z < 1.08 ) = 0.86
z = 1.08
Using z-score formula,
x = z * +
x = 1.08 * 14 + 120 = 135.12
X = 135.12
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