THE NORMAL DISTRIBUTION FUNCTION
If Z has a normal distribution with mean 0 and
variance 1, then, for each value of z, the table gives
the value of Φ(z), where
Φ(z) = P(Z ⩽ z).
For negative values of z, use Φ(–z) = 1 – Φ(z).
1 2 3 4 5 6 7 8 9
z 0 1 2 3 4 5 6 7 8 9
ADD
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 4 8 12 16 20 24 28 32 36
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 4 8 12 16 20 24 28 32 36
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 4 8 12 15 19 23 27 31 35
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 4 7 11 15 19 22 26 30 34
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 4 7 11 14 18 22 25 29 32
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 3 7 10 14 17 20 24 27 31
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 3 7 10 13 16 19 23 26 29
0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 3 6 9 12 15 18 21 24 27
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 3 5 8 11 14 16 19 22 25
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 3 5 8 10 13 15 18 20 23
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 2 5 7 9 12 14 16 19 21
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 2 4 6 8 10 12 14 16 18
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 2 4 6 7 9 11 13 15 17
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 2 3 5 6 8 10 11 13 14
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1 3 4 6 7 8 10 11 13
1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1 2 4 5 6 7 8 10 11
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1 2 3 4 5 6 7 8 9
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1 2 3 4 4 5 6 7 8
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1 1 2 3 4 4 5 6 6
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 1 1 2 2 3 4 4 5 5
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 0 1 1 2 2 3 3 4 4
2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 0 1 1 2 2 2 3 3 4
2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 0 1 1 1 2 2 2 3 3
2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 0 1 1 1 1 2 2 2 2
2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 0 0 1 1 1 1 1 2 2
2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 0 0 0 1 1 1 1 1 1
2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 0 0 0 0 1 1 1 1 1
2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 0 0 0 0 0 1 1 1 1
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 0 0 0 0 0 0 0 1 1
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986 0 0 0 0 0 0 0 0 0
Critical values for the normal distribution
If Z has a normal distribution with mean 0 and
variance 1, then, for each value of p, the table
gives the value of z such that
P(Z ⩽ z) = p.
p 0.75 0.90 0.95 0.975 0.99 0.995 0.9975 0.999 0.9995
z 0.674 1.282 1.645 1.960 2.326 2.576 2.807 3.090 3.291
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