Probability of a standard normal distribution.

Hi! I have 2 problems on the standard normal distribution. Please help me with this.

  1. For a standard normal distribution, F(0) is 0.5.

  2. For the standard normal distribution, P(0 <= z <=1.96) is 0.4750.


  1. Need clarification on the reason: because z- value of 0 gives has only a 5% occurence?

  2. 2 tailed test with a z value of 1.96 has a probability of 97.5% occurence. But since it is only 0 to 1.96, isn’t it suppose to be 48.75%(=97.5/2)?

  1. F(0) means P(Z ≤ 0)
  2. The 97.5% number you quote is that F(1.96) = P(Z ≤ 1.96) = 97.5%. So P(0 ≤ Z ≤ 1.96) = P(Z ≤ 1.96) − P(Z ≤ 0) = 97.5% − 50% = 47.5%.
  1. This has to do with the positive and negative infinity of the distribution on both tails. F(0) sums from negative infinitive to zero and gives you 0.5. Remember: The sum on both sides must be 1. In contrast, when you take a z-value of 1.96 you cut off the infinite part of the distribution which leaves you with a probability lower than 0.5 (0.475 in this case).

  2. If you have a 2-tailed test you test on both sides of the distribution (you want to know whether the population mean is higher or lower of the sample mean), hence it is not correct to take the half of the 97.5%.



Thanks. That makes a lot of sense. Why didn’t I think of that in the first place (Bangs head).

Btw, how you do you make that greater than or equal symbol?

I forgot about z-distribution being 2 tailed the moment I saw 0. Thanks for your help.

I copy and paste from Word. I have to use ctrl-v to paste.

You’re quite welcome.