Question

4. Given that z is a standard normal random variable, compute the following probabilities.

1. P(z ≤ -0.71)
2. P(z ≤ 1.82)
3. P(z ≥ -0.71)
4. P(z ≥ 1.22)
5. P( –1.71 ≤ z ≤ 2.88)
6. P( 0.56 ≤ z ≤ 1.07)
7. P( –1.65 ≤ z ≤ –1.65)

5. Given that z is a standard normal random variable, find z, for each situation.

1. The area to the left of z is 0.9608
2. The area to the right of z is .0102
3. The area between o and z is 0.4750

Answer 1

This is a normal distribution question with


a)
This implies that
P(z < -0.71) = 0.2389

b)
This implies that
P(z < 1.82) = 0.9656

c)
This implies that
P(z > -0.71) = 0.7611

d)
This implies that
P(z > 1.22) = 0.1112

e)
This implies that
P(-1.71 < z < 2.88) = P(z < z2) - P(z < z1)
P(-1.71 < z < 2.88) = 0.9980116241451057 - 0.9980116241451057
P(-1.71 < z < 2.88) = 0.9544

f)
This implies that
P(0.56 < z < 1.07) = P(z < z2) - P(z < z1)
P(0.56 < z < 1.07) = 0.8576903456440608 - 0.8576903456440608
P(0.56 < z < 1.07) = 0.1454

g)
This implies that
P(-1.65 < z < -1.65) = P(z < z2) - P(z < z1)
P(-1.65 < z < -1.65) = 0.0494714680336481 - 0.0494714680336481
P(-1.65 < z < -1.65) = 0.0


h) Given in the question, p = 0.9608
P(X < x) = 0.9608
This implies that
P(Z < 1.7600460440337644) = 0.9608
z = 1.7600460440337644


i) Given in the question, right of z = 0.0102 means left of z = 0.9898
P(X < x) = 0.9898
This implies that
P(Z < 2.3189084659001677) = 0.9898
z = 2.3189084659001677

j)
Given in the question, area of 0.475 between 0 and z
area to left of 0 is 0.5 so, the area to ledt of z must be 0.5+0.475 = 0.975

P(X < x) = 0.975
This implies that
P(Z < 1.959963984540054) = 0.975
z = 1.959963984540054

PS: you have to refer z score table to find the final probabilities.

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