This document provides information about the standard normal distribution and calculates probabilities for various z-scores. It gives the probability that a z-score will fall within certain ranges, as well as finding z-scores that correspond to specific probabilities. For example, it finds that the z-score corresponding to a 0.4207 probability above and below the mean is 0.2.

1. Normal Distribution
GiventhatZ isthe standard normal distribution.Findthe valuesforthe following
1 P ( Z> 0.2)
P ( Z< -0.2)
0.42074
6 P( -1.5 < Z< -0.6)
P ( Z > -1.5) – P ( Z > -0.6)
P ( Z< 1.5) – P ( Z <0.6)
0.93319- 0.72575
0.2075
11 P ( |z|≥1.334)
P ( z ≥1.334) + p ( -z≥ 1.334)
P ( z < −1.334) + 𝑃 ( 𝑧 <
−1.334)
0.0912 + 0.0912
0.1824
2 P ( Z <- 0.6)= 0.2743
{Mode SD 1 Shift3
P 1 ( -0.6)}
7 P ( 0 < Z < 1.511)
P ( Z > 0 ) – P ( Z > 1.511)
P ( Z < 0 ) – P ( Z < -1.511)
0.5-0.006539
0.4934
P (|z| ≤ 1.112)
1- P ( |Z|≥1.112)
1- [ P ( Z ≥ 1.112 ) + 𝑃 ( 𝑍 ≥
1.112 ) ]
1- [ P ( Z < -1.112) + P ( Z <- 1.112)]
1- [ 0.13307 + 0.13307]
1- 0.26614
0.7338
3 P ( Z >- 1.511)
P ( Z < 1.511)
[ we haveto change> to
< to find Z fromP 1
Calculator]
0.9346
8 P ( -1.013 <Z< -0.203)
P ( Z > -1.013 ) – P ( Z > -
0.203)
P ( Z < 1.013) – P ( Z <
0.203)
0.84447-0.58043
0.2640
4 P(Z<1.327)
0.9077
9 P ( -0.203 ≤ z ≤ 1.327)
P ( z > -0.203) – p ( z >
1.327)
P ( z < 0.203) – p ( z < -
1.327)
0.58043- 0.09225
0.48818
0.4882
5 P(0.2 < Z < 1.2)
P ( Z>0.2) – P ( Z >1.2)
P ( Z < - 0.2) – P ( Z < -
1.2)
0.42074 -0.11507
0.3056
10 P ( z < 0.549)
0.7085

2. Findthe z score foreach
of the following
P ( Z> z ) = 0.4207
P ( Z <- z ) = 0.4207
Z = 0.2
P ( Z > z ) = 0.1151
P ( Z <-z) = 0.1151
Z= 1.2
P(Z<z)=0.2743
Z= -0.6