Let \(Z\) be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate.
a. \(P(0 \leq Z \leq 2.17)\)
b. \(P(0 \leq Z \leq 1)\)
c. \(P(-2.50 \leq Z \leq 0)\)
d. \(P(-2.50 \leq Z \leq 2.50)\)
e. \(P(Z \leq 1.37)\)
f. \(P(-1.75 \leq Z)\)
g. \(P(-1.50 \leq Z \leq 2.00)\)
h. \(P(1.37 \leq Z \leq 2.50)\)
i. \(P(1.50 \leq Z)\)
j. \(P(|Z| \leq 2.50)\)
By using standard normal distribution table,
a.
\(P(0 \leq Z \leq 2.17)=P(Z \leq 2.17)-P(\leq 0)=\Phi(2.17)-\Phi(0)\)
\(=0.9850-0.5=0.4850\)
b.
\(P(0 \leq Z \leq 1)=P(Z \leq 1)-P(\leq 0)=\Phi(1)-\Phi(0)\)
\(=0.8413-0.5=0.3413\)
c.
\(P(-2.50 \leq Z \leq 0)=P(Z \leq 0)-P(\leq-2.50)=\Phi(0)-\Phi(-2.50)\)
\(=\Phi(0)-[1-\Phi(2.50)]=0.5-1+0.9938=0.4938\)
d.
\(P(-2.50 \leq Z \leq 2.50)=P(Z \leq 2.50)-P(\leq-2.50)=\Phi(2.50)-\Phi(-2.50)\)
\(=\Phi(2.50)-[1-\Phi(2.50)]=2 * 0.9938-1=0.9876\)
e.
\(P(Z \leq 1.37)=\Phi(1.37)=0.9147\)
\(\mathrm{f}\)
\(P(-1.75 \leq Z)=P(Z \geq-1.75)=1-P(Z<-1.75)=1-\Phi(-1.75)\)
\(=1-[1-\Phi(1.75)]=\Phi(1.75)=0.9599\)
\(\mathrm{g}\)
\(P(-1.50 \leq Z \leq 2.00)=P(Z \leq 2.00)-P(\leq-1.50)=\Phi(2.00)-\Phi(-1.50)\)
\(=\Phi(2.00)-[1-\Phi(1.50)]=0.9772-1+0.9332=0.9104\)
\(\mathrm{h}\)
\(P(1.37 \leq Z \leq 2.50)=P(Z \leq 2.50)-P(\leq 1.37)=\Phi(2.50)-\Phi(1.37)\)
\(=0.9938-0.9147=0.0791\)
\(\mathbf{i}\)
\(P(1.50 \leq Z)=P(Z \geq 1.50)=1-P(Z<1.50)=1-\Phi(1.50)\)
\(=1-0.9332=0.0668\)
i.
\(P(|Z| \leq 2.50)=P(-2.50 \leq Z \leq 2.50)\)
\(=P(Z \leq 2.50)-P(Z \leq-2.50)=\Phi(2.50)-\Phi(-2.50)\)
\(=\Phi(2.50)-[1-\Phi(2.50)]=2 * \Phi(2.50-1)=2 * 0.9938-1=0.9876\)
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