91影视

In both directions, the standard normal curve extends endlessly, approaching but never reaching the horizontal axis. The bell-shaped standard normal curve is centered at \({\rm{z = 0}}\). Between \({\rm{z = - 3}}\)and \({\rm{z = 3}}\), almost all of the area under the standard normal curve is found.

No, it is necessary to tabulate the \(\phi (z)\)for \({\rm{x}}\) negative. Using the symmetry of the standard normal distribution curve about\({\rm{0}}\), we know that area to the left of \({\rm{z = - }}{{\rm{z}}_{\rm{0}}}\)is equal to the area to the right of\({\rm{z = }}{{\rm{z}}_{\rm{0}}}\). Hence, we can write:

\(\phi \left( { - {z_0}} \right) = 1 - \phi \left( {{z_0}} \right)\)Step 2

(a) We have to calculate \(P( - 1.72 \le Z \le - 0.55)\)

\(\begin{aligned}P( - 1.72 \le Z \le - 0.55) &= \phi ( - 0.55) - \phi ( - 1.72)\\ &= (1 - \phi (0.55)) - (1 - \phi (1.72))\\ &= (1 - 0.7088) - (1 - 0.9573)\\ &= 0.2912 - 0.0427P( - 1.72 \le Z \le - 0.55)\\ &= 0.2485\end{aligned}\)

Proposition: Let \({\rm{Z}}\)be a continuous \({\rm{rv}}\) with \(cdf\phi (z)\). Then for any two numbers a and \({\rm{b}}\)with \({\rm{a < b,}}\)

\(P(a \le Z \le b) = \phi (b) - \phi (a)\)

Therefore, the result is \(P( - 1.72 \le Z \le - 0.55) = 0.2485\).

b)

We have to calculate \(P( - 1.72 \le Z \le 0.55)\)

\(\begin{aligned}P( - 1.72 \le Z \le 0.55) &= \phi (0.55) - \phi ( - 1.72)\\ &= \phi (0.55) - (1 - \phi (1.72))\\ &= 0.7088 - (1 - 0.9573)\\ &= 0.7088 - 0.0427P( - 1.72 \le Z \le 0.55)\\ &= 0.6661\end{aligned}\)

Therefore, the result is \(P( - 1.72 \le Z \le 0.55) = 0.6661\).