Given that z is a standard normal random variable compute th
Solution
a)
Using a table/technology, the left tailed area of this is
P(z < -1 ) = 0.158655254 [ANSWER]
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b)
Using a table/technology, the right tailed area of this is
P(z > -1 ) = 0.841344746 [ANSWER]
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c)
Using a table/technology, the right tailed area of this is
P(z > -1.5 ) = 0.933192799 [ANSWER]
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d)
Using a table/technology, the right tailed area of this is
P(z > -2.5 ) = 0.993790335 [ANSWER]
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e)
z1 = lower z score = -3
z2 = upper z score = 0
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.001349898
P(z < z2) = 0.5
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.498650102 [ANSWER]
