Answer:
a. Using a standard normal table or calculator, we find that z = -1.28 satisfies P(z≤zo) = 0.0981.
b. Since the standard normal distribution is symmetric, P(-Zo≤z≤zo) = 0.99 is equivalent to P(z≤-zo) = 0.005. Using a standard normal table or calculator, we find that z = -2.33 satisfies this probability.
c. Since the standard normal distribution is symmetric, P(-Z₁ ≤z≤zo) = 0.95 is equivalent to P(0 ≤z≤Zo) = 0.475. Using a standard normal table or calculator, we find that z = 1.96 satisfies this probability.
d. Since the standard normal distribution is symmetric, P(-Z₁≤z≤z) = 0.8994 is equivalent to P(0≤z≤Z₁) = 0.4497. Using a standard normal table or calculator, we find that z = 2.66 satisfies this probability.
e. Since the standard normal distribution is symmetric, P(-Zo≤z≤0) = 0.5 - P(0≤z≤Zo) = 0.5 - 0.3106 = 0.1894. Using a standard normal table or calculator, we find that z = -0.84 satisfies this probability.
f. Since the standard normal distribution is symmetric, P(-3≤z≤3) = 0.998. Therefore, P(z>3 or z<-3) = 0.002.
g. P(Z<z) = 0.0014 is equivalent to P(z>-z₁) = 0.0014, where z₁ is the z-value such that P(z≤z₁) = 0.0014. Using a standard normal table or calculator, we find that z₁ = -2.96. Therefore, z > 2.96 satisfies P(Z<z) = 0.0014.
h. P(z≤z) = 0.5 + 0.0014/2 = 0.5007. Using a standard normal table or calculator, we find that z = 2.59 satisfies this probability.
Explanation: