Suppose that $Z \sim N(0, 1)$ . Find:
$\begin{align} &\mathbf{(a)}\, P(Z \le −0.77)\\ &\mathbf{(b)}\, P(Z \ge 0.32)\\ &\mathbf{(c)}\, P(−3.09 \le Z \le −1.59)\\ &\mathbf{(d)}\, P(−0.82 \le Z \le 1.80)\\ &\mathbf{(e)}\, P(|Z| \ge 0.91)\\ &\mathbf{(f)}\,\text{The value of } x \text{ for which } P(Z \le x) = 0.23\\ &\mathbf{(g)}\,\text{The value of } x \text{ for which } P(Z \ge x) = 0.51\\ &\mathbf{(h)}\,\text{The value of } x \text{ for which } P(|Z| \ge x) = 0.42\\ \end{align}$

Solution

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Step 1

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Assume that the random variable $Z$ has $\textit{the standard normal distribution}$ :

\color{#c34632}{Z \sim N(0,1)}

Hence, $E(Z)=\mu=0$ and $\mathrm{Var}(Z)=\sigma^2=1$ . Also, in this case, the probability density function of $Z$ is given by

\phi(z)=\dfrac{1}{\sqrt{2\pi}}e^{-z^2/2}\,\,,\,\,-\infty \le z \le \infty

and its cumulative distribution function can be calculated from the expression

\Phi(z)=P(Z \le z)=\int\limits_{-\infty}^z \dfrac{1}{\sqrt{2\pi}}e^{-t^2/2} dt

We will calculate some probabilities by using $\color{#c34632}{\textbf{Table I}}$ at the end of the book:

$\colorbox{Apricot}{\textbf{(a)}}$ Figure1 below illustrates the part of Table I in which the value of $\Phi(-0.77)$ is located. On the right is illustrated function $\phi(z)$ , where marked area represents the probability $P(Z \le -0.77)=\Phi(-0.77)$ . Hence,

P(Z \le -0.77)=\Phi(-0.77)=\boxed{\bf{0.2206}}

$\colorbox{Apricot}{\textbf{(b)}}$ The probability $P(Z \ge 0.32)$ can be obtained in the following way:

\begin{align*} P(Z \ge 0.32)&=1-P(Z < 0.32)=1-[P(Z < 0.32)+\underbrace{P(Z = 0.32)}_{=0}]\\ &=1-P(Z \le 0.32)=1-\Phi(0.32)\\ &=1-0.6255=\boxed{\bf{0.3745}} \end{align*}

This is illustrated in Figure2. Look at Table I and notice: $0.3745=\Phi(-0.32)$ . Hence, $1-\Phi(0.32)=\Phi(-0.32)$ . Becouse of the symmetry of the standard normal distribution about 0, this relationship holds in general. So, we have: