# Calculate the critical value of the standard Normal distribution associated with a confidence level of 96.4% given that only tabulated values for the 96th and 97th percentiles of the standard...

Calculate the critical value of the standard Normal distribution associated with a confidence level of 96.4% given that only tabulated values for the 96th and 97th percentiles of the standard Normal distribution are available.

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### 1 Answer

Given the critical values for 96th and 97th percentiles

`Phi^(-1)(0.96) = z_(0.96) = 1.751` and

`Phi^(-1)(0.97) = z_(0.97) = 1.881`

use *'linear interpolation'* to find the 96.4th percentile `Phi^(-1)(0.964) = z_(0.964)`

The formula for linear interpolation is as follows:

`x = ((x_1-x)x_0 + (x-x_0)x_1)/((x_1-x_0))`

where `x` is the value (between two known values `x_0` and `x_1`, `x_1>x_0` ) that we wish to find. The value of `(x_1-x_0)` is the distance between the two known values and `(x_1-x)` and `(x-x_0)` are the distances respectively between `x` and `x_1` and `x` and `x_0` . If `x` is closer to `x_0` it's value should be closer to that of `x_0` than that of `x_1` and vice versa. In fact the weight that the value `x_0` carries in the *'weighted average'* of `x_0` and `x_1` is equal to `((x_1-x))/((x_1-x_0))` , that is, the distance between `x` and `x_1` over the total distance.

Now, 0.964 is 4/10ths of the distance between 0.96 and 0.97 so that our estimate of 0.964 will be 6/10 times 0.96 (`x_0`) and 4/10 times 0.97 `(x_1)`. This makes sense as 0.964 is closer to `x_0` than `x_1`. The interpolation is linear because it is equivalent to drawing a straight line between the values at `x_0` and at `x_1` and reading off the value at the point `x` between them (try this for yourself!).

**Answer: Therefore, the quantile at the 96.4th percentile of the standard Normal can be linearly interpolated to be **

**(6/10 x 1.751) + (4/10 x 1.881) = 1.80 to 3sf (this is in fact totally accurate to 3sf)**