To find the values of the standard normal random variable z for the given probabilities, we can use a standard normal distribution table or calculator. A. P(z≤z0)=0.6997, z0=0.52. B. P(−z0≤z≤z0)=0.8026, z0=1.29. C. P(−z0≤z≤0)=0.4702, z0=-0.63. D. P(−1<z<z0)=0.159, z0=0.49
A. P(z≤z0)=0.6997
To find the value of z0 such that P(z≤z0) = 0.6997, we can use a standard normal distribution table or calculator. Looking up the value, we get z0 ≈ 0.52.
B. P(−z0≤z≤z0)=0.8026
To find the value of z0 such that P(−z0≤z≤z0) = 0.8026, we can use the symmetry property of the standard normal distribution. Since the probability is split equally on both sides of the mean, we can find the z-score that corresponds to a probability of (1-0.8026)/2 = 0.0987 on the left tail. Looking up this value, we get z ≈ -1.29. Therefore, z0 ≈ 1.29.
C. P(−z0≤z≤0)=0.4702
To find the value of z0 such that P(−z0≤z≤0) = 0.4702, we can use the symmetry property again. Since the probability is split equally on both sides of the mean, we can find the z-score that corresponds to a probability of (1-0.4702)/2 = 0.2649 on the right tail. Looking up this value, we get z ≈ 0.63. Therefore, z0 ≈ -0.63.
D. P(−1<z<z0)=0.159
To find the value of z0 such that P(−1<z<z0) = 0.159, we can subtract the probability of the left tail (P(z<-1)) from the total probability (0.159 + P(z<-1) = 1). Looking up the value of P(z<-1) in a standard normal distribution table, we get 0.1587. Therefore, the probability of the right tail (P(z<z0)) is 1 - 0.159 - 0.1587 = 0.6823. Looking up the z-score that corresponds to this probability, we get z0 ≈ 0.49.
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