Question

find the following probabilities for the standard normal random variable z:
(a) ( p(-1.56 leq z leq 1.3)=)
(b) ( p(-1.88 leq z leq 1.02)=)
(c) ( p(z geq 2.08)=)
(d) ( p(z>-1.69)=)

Explanation:

Step1: Recall the formula for \(P(a\leq z\leq b)\)

\(P(a\leq z\leq b)=\Phi(b)-\Phi(a)\), where \(\Phi(z)\) is the cumulative - distribution function of the standard normal distribution. For \(P(z\geq c) = 1-\Phi(c)\) and \(P(z > d)=1 - \Phi(d)\) (since for a continuous distribution \(P(z\geq d)=P(z > d)\)).

Step2: Use the standard - normal table (or a calculator with a normal - distribution function)

• (a)
• For \(z = 1.3\), from the standard - normal table, \(\Phi(1.3)=0.9032\)
• For \(z=-1.56\), \(\Phi(-1.56)=1 - \Phi(1.56)\). Since \(\Phi(1.56)=0.9406\), then \(\Phi(-1.56)=1 - 0.9406=0.0594\)
• \(P(-1.56\leq z\leq1.3)=\Phi(1.3)-\Phi(-1.56)=0.9032-(1 - 0.9406)=0.9032 - 0.0594=0.8438\)
• (b)
• For \(z = 1.02\), \(\Phi(1.02)=0.8461\)
• For \(z=-1.88\), \(\Phi(-1.88)=1 - \Phi(1.88)\). Since \(\Phi(1.88)=0.9699\), then \(\Phi(-1.88)=1 - 0.9699 = 0.0301\)
• \(P(-1.88\leq z\leq1.02)=\Phi(1.02)-\Phi(-1.88)=0.8461-(1 - 0.9699)=0.8461 - 0.0301=0.816\)
• (c)
• For \(z = 2.08\), \(\Phi(2.08)=0.9812\)
• \(P(z\geq2.08)=1-\Phi(2.08)=1 - 0.9812=0.0188\)
• (d)
• For \(z=-1.69\), \(\Phi(-1.69)=1 - \Phi(1.69)\). Since \(\Phi(1.69)=0.9545\), then \(\Phi(-1.69)=1 - 0.9545=0.0455\)
• \(P(z > - 1.69)=1-\Phi(-1.69)=\Phi(1.69)=0.9545\)

Answer:

(a) \(0.8438\)
(b) \(0.816\)
(c) \(0.0188\)
(d) \(0.9545\)