Homework SourseA recent study of the wages of cable car operators showed th
A recent study of the wages of cable car operators showed that the mean annual income were normally distributed with a mean income of 33,000 with a deviation of 4000. If we selected an operator at random
Fiver percent earned how much or more
What percent earned between 35,000 and 36,500
What percent earned between 24,00 and 37,000
Solution
a)
First, we get the z score from the given left tailed area. As
Left tailed area = 1 - 0.05 = 0.95
Then, using table or technology,
z = 1.644853627
As x = u + z * s,
where
u = mean = 33000
z = the critical z score = 1.644853627
s = standard deviation = 4000
Then
x = critical value = 39579.41451 [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 = 35000
x2 = upper bound = 36500
u = mean = 33000
s = standard deviation = 4000
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = 0.5
z2 = upper z score = (x2 - u) / s = 0.875
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.691462461
P(z < z2) = 0.809213047
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.117750586 [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 = 24000
x2 = upper bound = 37000
u = mean = 33000
s = standard deviation = 4000
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2.25
z2 = upper z score = (x2 - u) / s = 1
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.012224473
P(z < z2) = 0.841344746
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.829120273 [ANSWER]
