Question

suppose z follows the standard normal distribution. calculate the following probabilities using the least three decimal places.
(a) ( p(z leq 1.77) )
(b) ( p(z > -0.99) )
(c) ( p(0.59 < z < 1.70) )

Explanation:

Step1: Find \( P(Z\leq1.77) \)

Using the standard - normal distribution table (or a calculator with a normal - distribution function, e.g., in Excel: NORM.S.DIST(1.77,TRUE)), we know that for a standard normal random variable \( Z\), \(P(Z\leq z)\) gives the cumulative probability.
\(P(Z\leq1.77) = 0.9616\)

Step2: Find \( P(Z > - 0.99)\)

Use the property \(P(Z>z)=1 - P(Z\leq z)\).
So \(P(Z > - 0.99)=1 - P(Z\leq - 0.99)\).
From the standard - normal table (or NORM.S.DIST(-0.99,TRUE) in Excel), \(P(Z\leq - 0.99)=0.1611\).
Then \(P(Z > - 0.99)=1 - 0.1611=0.8389\)

Step3: Find \(P(0.59 < Z < 1.70)\)

Use the property \(P(a < Z < b)=P(Z < b)-P(Z < a)\)
From the standard - normal table (or NORM.S.DIST(1.70,TRUE) and NORM.S.DIST(0.59,TRUE) in Excel), \(P(Z < 1.70)=0.9554\) and \(P(Z < 0.59)=0.7224\)
\(P(0.59 < Z < 1.70)=0.9554 - 0.7224=0.233\)

Answer:

(a) \(0.962\)
(b) \(0.839\)
(c) \(0.233\)