Question

find probabilities for the standard and non - standard normal distribution.
you may round your answers to four decimal places
a. ( p(z>2.08)=)
b. ( p(z < - 2.07)=)
c. ( p(-0.75 < z < 1.12)=)
d. if ( mu = 77 ) and ( sigma = 14 ), find ( p(51 < x < 91)=)

Explanation:

Step1: Find \(P(z > 2.08)\)

Using the standard normal table, \(P(z\leq2.08) = 0.9812\). Then \(P(z > 2.08)=1 - P(z\leq2.08)\).

$$P(z > 2.08)=1 - 0.9812=0.0188$$

Step2: Find \(P(z < - 2.07)\)

Using the standard normal table, \(P(z < - 2.07)=P(z>2.07)\) (by symmetry of the standard normal distribution). \(P(z\leq2.07) = 0.9808\), so \(P(z>2.07)=1 - 0.9808 = 0.0192\). Thus \(P(z < - 2.07)=0.0192\)

Step3: Find \(P(-0.75 < z < 1.12)\)

Using the standard normal table, \(P(z < 1.12)=0.8686\) and \(P(z < - 0.75)=0.2266\). Then \(P(-0.75 < z < 1.12)=P(z < 1.12)-P(z < - 0.75)\)

$$P(-0.75 < z < 1.12)=0.8686 - 0.2266=0.6420$$

Step4: Find \(P(51 < x < 91)\) with \(\mu = 77\) and \(\sigma = 14\)

First, standardize the values. For \(x = 51\), \(z_1=\frac{51 - 77}{14}=\frac{-26}{14}\approx - 1.86\). For \(x = 91\), \(z_2=\frac{91 - 77}{14}=\frac{14}{14}=1\)
Using the standard normal table, \(P(z < 1)=0.8413\) and \(P(z < - 1.86)=0.0314\)
Then \(P(51 < x < 91)=P(-1.86 < z < 1)=P(z < 1)-P(z < - 1.86)\)

$$P(51 < x < 91)=0.8413-0.0314 = 0.8099$$

Answer:

a. \(0.0188\)
b. \(0.0192\)
c. \(0.6420\)
d. \(0.8099\)