Homework Soursesuppose x is a normally distributed random variable with mea
suppose x is a normally distributed random variable with mean 16 and Standard D 2. Find each of the following probabilities.
Solution
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 19
u = mean = 16
s = standard deviation = 2
Thus,
z = (x - u) / s = 1.5
Thus, using a table/technology, the right tailed area of this is
P(z > 1.5 ) = 0.066807201 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 15.5
u = mean = 16
s = standard deviation = 2
Thus,
z = (x - u) / s = -0.25
Thus, using a table/technology, the left tailed area of this is
P(z < -0.25 ) = 0.401293674 [ANSWER]
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c)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 17.52
x2 = upper bound = 20.86
u = mean = 16
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0.76
z2 = upper z score = (x2 - u) / s = 2.43
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.776372708
P(z < z2) = 0.992450589
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.216077881 [ANSWER]
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d)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 11.5
x2 = upper bound = 19.54
u = mean = 16
s = standard deviation = 2
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2.25
z2 = upper z score = (x2 - u) / s = 1.77
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.012224473
P(z < z2) = 0.96163643
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.949411957 [ANSWER]
