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# 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*}

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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:

$(\star)\,\,\,\Phi(z)+\Phi(-z)=1$

Using general result $\color{#4257b2}{P(a \le Z \le b)=\Phi(b)-\Phi(a)}$ , we obtain :

$\colorbox{Apricot}{\textbf{(c)}}\,\,\,P(-3.09 \le Z \le -1.59)=\Phi(-1.59)-\Phi(-3.09)=0.0559-0.0010=\boxed{\bf{0.0549}}$

$\colorbox{Apricot}{\textbf{(d)}}\,\,\,P(-0.82 \le Z \le 1.80)=\Phi(1.80)-\Phi(-0.82)=0.9641-0.2061=\boxed{\bf{0.758}}$

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