Homework SourseLet X be normally distributed with mean mu 102 and standard
Solution
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 100
u = mean = 102
s = standard deviation = 34
Thus,
z = (x - u) / s = -0.06
Thus, using a table/technology, the left tailed area of this is
P(z < -0.06 ) = 0.476077817 [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 95
x2 = upper bound = 110
u = mean = 102
s = standard deviation = 34
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.21
z2 = upper z score = (x2 - u) / s = 0.24
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.416833837
P(z < z2) = 0.594834872
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.178001035 [ANSWER]
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c)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.063
Then, using table or technology,
z = -1.530067588
As x = u + z * s,
where
u = mean = 102
z = the critical z score = -1.53
s = standard deviation = 34
Then
x = critical value = 49.98 [ANSWER]
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D)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.648
Then, using table or technology,
z = 0.379926467
As x = u + z * s,
where
u = mean = 102
z = the critical z score = 0.38
s = standard deviation = 34
Then
x = critical value = 114.92 [ANSWER]
