# Normal distribution

Mathsand Stat / Maths Module 1 - Normal distribution I amasked to find the value of a and b if P(0 <Z < a) = 0.4990 and P(0 < Z< b) = 0.49905, where Z ~ N(0, 1) Here isthe given standard normal table: 3.08 ... 3.10 3.11 ... 3.13 4990 ... 4990 4991 ... 4991 I am notsure if the value of a is 3.08 or 3.09or 3.10.Ans, I am not sure if the value of b is 3.095 or 3.10 or 3.105 or 3.11 or 3.115or 3.12 or 3.13. Pleaseanswer me. (I know some technique on linear interpolation.) Thanks.

### 1 Answer

- 良師益友Lv 78 years agoFavorite Answer
My analysis as follows:

1) From the normal dist. curve and its corresponding table, we can see that either 3.08, 3.09 or 3.10 can all give the prob. 0.4990.

In such a case, we should adopt the "Mean" value of all possible values here, which means, the value of "a"

= ( 3.08 + 3.09 + 3.10 ) / 3

= 3.09

The concept of Interpolation has been incorporated. It is actually controversial as all three mentioned values can fulfill the requirement of questions. But if we realize those concepts above rounding off ( and limits ) we realize we must select the central data for minimum error.

2) First, the value must lie between 3.10 and 3.11 by considering their possible true values after rounding off. We should realize than for z = 3.10, its prob. must be less than 0.4995. Otherwise, it has been rounded up to 0.4991.

Once again, we choose the mean of these two data.

Hence, the value of b

= ( 3.10 + 3.11 ) / 2

= 3.105

Such questions touching the extreme values would appear in Math. & Stat. and it is possible to be seen again in Module 1. Such type of questions should be practiced more for Module 1 students, in my opinion.

Hope I can help you.

Source(s): Mathematics Teacher Mr. Ip