584 For a normal random variable X with a mean of 8 and stan
Solution
First, we get the z score from the given left tailed area. As
Left tailed area = 0.2
Then, using table or technology,
z = -0.841621234
As x = u + z * s / sqrt(n)
where
u = mean = 8
z = the critical z score = -0.841621234
s = standard deviation = 2
n = sample size = 1
Then
x = critical value = 6.316757533 [ANSWER, A]
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B.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.72
Then, using table or technology,
z = 0.582841507
As x = u + z * s / sqrt(n)
where
u = mean = 8
z = the critical z score = 0.582841507
s = standard deviation = 2
n = sample size = 1
Then
x = critical value = 9.165683015 [ANSWER, B]
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C.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.95
Then, using table or technology,
z = 1.644853627
As x = u + z * s / sqrt(n)
where
u = mean = 8
z = the critical z score = 1.644853627
s = standard deviation = 2
n = sample size = 1
Then
x = critical value = 11.28970725 [ANSWER, C]
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D.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.05
Then, using table or technology,
z = -1.644853627
As x = u + z * s / sqrt(n)
where
u = mean = 8
z = the critical z score = -1.644853627
s = standard deviation = 2
n = sample size = 1
Then
x = critical value = 4.710292746 [ANSWER, D]
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