Answer:
Approximately [tex]0.428[/tex].
Step-by-step explanation:
(This approach requires looking up values on a [tex]z[/tex]-table. Calculators aren't required.)
A typical [tex]z[/tex]-table lists probabilities in the form [tex]P(Z \le z)[/tex] (or equivalently, [tex]P(Z < z)[/tex],) where [tex]z\![/tex] is real number (typically greater than or equal to [tex]0[/tex].)
However, the question asks for a probability in the form [tex]P(a < Z < b)[/tex]. It would be necessary to rewrite this probability for this [tex]\!z[/tex]-table approach to work.
Let [tex]a[/tex] and [tex]b[/tex] denote two real numbers, with [tex]b > a[/tex]. [tex]P(a < Z < b)[/tex] would be equal to [tex]P(Z < b) - P(Z < a)[/tex]. In this question:
[tex]\begin{aligned} & P(0 < Z < 1.46) \\ & = P(Z < 1.46) - P(Z < 0) \end{aligned}[/tex].
Look up the value of [tex]P(Z < 1.46)[/tex] on a [tex]z[/tex]-table: [tex]P(Z < 1.46) \approx 0.927855[/tex].
On the other hand, [tex]P(Z < 0) = 0.5[/tex] because the Gaussian distribution is symmetric.
Therefore:
[tex]\begin{aligned} & P(0 < Z < 1.46) \\ & = P(Z < 1.46) - P(Z < 0) \\ &\approx 0.927855 - 0.5 \\ &\approx 0.428\end{aligned}[/tex].
