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Chapter 5: Q2E (page 315)

Suppose that X has the normal distribution for which the mean is 1 and the variance is 4. Find the value of each of the following probabilities:

(a). \({\rm P}\left( {X \le 3} \right)\)

(b). \({\rm P}\left( {X > 1.5} \right)\)

(c). \({\rm P}\left( {X = 1} \right)\)

(d). \({\rm P}\left( {2 < X < 5} \right)\)

(e). \({\rm P}\left( {X \ge 0} \right)\)

(f). \({\rm P}\left( { - 1 < X < 0.5} \right)\)

(g). \({\rm P}\left( {\left| X \right| \le 2} \right)\)

(h). \({\rm P}\left( {1 \le - 2X + 3 \le 8} \right)\)

Short Answer

Expert verified

(a). Value of \({\rm P}\left( {X \le 3} \right)\) is 0.8413

(b). Value of \({\rm P}\left( {X > 1.5} \right)\) is 0.4013

(c). Value of \({\rm P}\left( {X = 1} \right) = 0\)

(d). Value of \({\rm P}\left( {2 < X < 5} \right)\)is \(0.2858\)

(e). Value of \({\rm P}\left( {X \ge 0} \right)\)is 0.6915

(f). Value ofP( -1 <X< 0.5) is 0.2426.

(g). Value of \({\rm P}\left( {\left| X \right| \le 2} \right)\)is 0.6247

(h). Value of \({\rm P}\left( {1 \le - 2X + 3 \le 8} \right)\)is 0.4599

Step by step solution

01

Given information.

X as the normal distribution with mean 1 and variance 4

02

Computing the probability.

Let \(Z = \frac{{\left( {X - 1} \right)}}{2}\)

Then Z has the standard normal distribution.

\({\rm P}\left( {X \le 3} \right) = {\rm P}\left( {Z \le 1} \right)\)

\(\begin{array}{c}{\rm P}\left( {X \le 3} \right) = \phi \left( 1 \right)\\ = 0.8413\end{array}\)

Value of\({\rm P}\left( {X \le 3} \right)\)is 0.8413

b.

\({\rm P}\left( {X > 1.5} \right) = {\rm P}\left( {Z > 0.25} \right)\)

\(\begin{array}{c}{\rm P}\left( {X > 1.5} \right) = 1 - \phi \left( {0.25} \right)\\ = 0.4013\end{array}\)

Value of\({\rm P}\left( {X > 1.5} \right)\)is 0.4013

c.

\({\rm P}\left( {X = 1} \right) = 0\)

Because X has a continuous distribution.

d.

\({\rm P}\left( {2 < X < 5} \right) = {\rm P}\left( {0.5 < Z < 2} \right)\)

\({\rm P}\left( {2 < X < 5} \right) = \phi \left( 2 \right) - \phi \left( {0.5} \right)\)

\({\rm P}\left( {2 < X < 5} \right) = 0.2858\)

Hence the value of\({\rm P}\left( {2 < X < 5} \right)\)is\(0.2858\)

e.

\(\begin{array}{c}{\rm P}\left( {X \ge 0} \right) = {\rm P}\left( {Z \ge - 0.5} \right)\\ = {\rm P}\left( {Z \le - 0.5} \right)\end{array}\)

\(\begin{array}{c}{\rm P}\left( {X \ge 0} \right) = \phi \left( {0.5} \right)\\ = 0.6915\end{array}\)

Hence ,\({\rm P}\left( {X \ge 0} \right)\)is 0.6915

f.

\[\begin{array}{c}{\rm P}\left( { - 1 < X < 0.5} \right) = {\rm P}\left( { - 1 < Z < - 0.25} \right)\\ = {\rm P}\left( { - 1 < Z < - 0.25} \right)\end{array}\]

\[\begin{array}{c}{\rm P}\left( { - 1 < X < 0.5} \right) = \phi \left( 1 \right) - \phi \left( {0.25} \right)\\ = 0.2426\end{array}\]

Hence the Value of P( -1 <X< 0.5) is 0.2426.

g.

\(\begin{array}{c}{\rm P}\left( {\left| X \right| \le 2} \right) = {\rm P}\left( { - 2 \le X \le 2} \right)\\ = {\rm P}\left( { - 1.5 \le Z \le 0.5} \right)\end{array}\)

\(\begin{array}{c}{\rm P}\left( {\left| X \right| \le 2} \right) = {\rm P}\left( {Z \le 0.5} \right) - {\rm P}\left( {Z \le 1.5} \right)\\ = {\rm P}\left( {Z \le 1.5} \right) - {\rm P}\left( {Z \ge 0.5} \right)\\ = \phi \left( {0.5} \right) - \left[ {1 - \phi \left( {1.5} \right)} \right]\end{array}\)

\({\rm P}\left( {\left| X \right| \le 2} \right) = 0.6247\)

Hence,\({\rm P}\left( {\left| X \right| \le 2} \right)\)is 0.6247

h.

\(\begin{array}{c}{\rm P}\left( {1 \le - 2X + 3 \le 8} \right) = {\rm P}\left( { - 2 \le - 2X \le 5} \right)\\ = {\rm P}\left( { - 2.5 \le X \le 1} \right)\end{array}\)

\(\begin{array}{c}{\rm P}\left( {1 \le - 2X + 3 \le 8} \right) = {\rm P}\left( { - 1.75 \le Z < 0} \right)\\ = {\rm P}\left( {0 \le Z < 1.75} \right)\end{array}\)

\(\begin{array}{c}{\rm P}\left( {1 \le - 2X + 3 \le 8} \right) = \phi \left( {1.75} \right) - \phi \left( 0 \right)\\ = 0.4599\end{array}\)

Hence \({\rm P}\left( {1 \le - 2X + 3 \le 8} \right)\)is 0.4599.

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