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radiance99 radiance99
wrote...
Posts: 20
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9 years ago
Let Z be a standard Normal random variable, i.e. Z ~N(0,1).
(a) Find Pr(Z < 1)
(b) Find Pr(Z> -1)
(c) Find Pr(-2<Z<1)
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wrote...
9 years ago
The N(0, 1) distribution is called the standard Normal distribution.

z    0.0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0    0.5000 5040 5080 5120 5160 5199 5239 5279 5319 5359
0.1   0.5398 5438 5478 5517 5557 5596 5636 5675 5714 5753
0.2    0.5793 5832 5871 5910 5948 5987 6026 6064 6103 6141
0.3    0.6179 6217 6255 6293 6331 6368 6406 6443 6480 6517
0.4    0.6554 6591 6628 6664 6700 6736 6772 6808 6844 6879
0.5    0.6915 6950 6985 7019 7054 7088 7123 7157 7190 7224
0.6   0.7257 7291 7324 7357 7389 7422 7454 7486 7517 7549
0.7    0.7580 7611 7642 7673 7704 7734 7764 7794 7823 7852
0.8    0.7881 7910 7939 7967 7995 8023 8051 8078 8106 8133
0.9    0.8159 8186 8212 8238 8264 8289 8315 8340 8365 8389
1.0    0.8413 8438 8461 8485 8508 8531 8554 8577 8599 8621

From this table we can identify that P (Z < 1.0) = 0.8413
If Z ∼ N(0, 1) what is P (Z > 0.92)?
We know that P(Z > 0.92) = 1−P(Z < 0.92) and we can calculate P (Z < 0.92) from the tables.
Thus, P (Z > 0.92) = 1 − 0.8212 = 0.1788

If Z ∼ N(0, 1) what is P (Z < −0.76)?
P(Z <−0.76)=P(Z >0.76) = 1−P(Z <0.76) = 1 − 0.7764
= 0.2236

If Z ∼ N(0, 1) what is P (−0.64 < Z < 0.43)?
P (Z < 0.43) − P (Z < −0.64) = 0.6664 − (1 − 0.7389)
= 0.4053
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