Let the random variable Z have a standard normal distribution. Fill in the blank
ID: 3151409 • Letter: L
Question
Let the random variable Z have a standard normal distribution.
Fill in the blanks. (Give your answers to four decimal places.)
a. P(Z 1.42 ) is .
b. P(Z <-1.33 ) is .
c. P(Z -2.00 Z 2.00) is .
d. P(Z >2.59 ) is .
e. P(0.13 <Z <2.44 ) is .
Let the random variable Z have a standard normal distribution.
Fill in the blanks. (Give your answers to four decimal places.)
a. P(Z 1.42 ) is .
b. P(Z <-1.33 ) is .
c. P(Z -2.00 Z 2.00) is .
d. P(Z >2.59 ) is .
e. P(0.13 <Z <2.44 ) is .
Explanation / Answer
A)
Using a table/technology, the left tailed area of this is
P(z < 1.42 ) = 0.9222 [ANSWER]
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b)
Using a table/technology, the left tailed area of this is
P(z < -1.33 ) = 0.0918 [ANSWER]
***************
c)
z1 = lower z score = -2
z2 = upper z score = 2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.0228
P(z < z2) = 0.9772
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.9544
Thus, those outside this interval is the complement = 0.0456 [ANSWER]
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d)
Using a table/technology, the left tailed area of this is
P(z < 2.59 ) = 0.9952
Thus, the right tailed area is the complement,
P(z > 2.59 ) = 0.0048 [ANSWER]
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e)
z1 = lower z score = 0.13
z2 = upper z score = 2.44
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.551716787
P(z < z2) = 0.992656369
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.440939582 [ANSWER]
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