Sir R. A. Fisher, a famous statistician, showed that the critical values of a chi-square distribution can be approximated by the standard normal distribution
$χ_{k}^{2} = \frac{{(z_{k} + \sqrt{2 v - 1})}^{2}}{2}$
where $v$ is the degrees of freedom and $z_{k}$ is the $z$ -score such that the area under the standard normal curve to the right of $z_{k}$ is $k$ . Use Fisher's approximation to find $χ_{0.975}^{2}$ and $χ_{0.025}^{2}$ with 100 degrees of freedom. Compare the results with those found in the table of the chi-square distribution.

In Fisher's approximation the value of $χ_{0.975}^{2}$ is $◻$ . (Type an integer or a decimal.)
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Answer

Round the result to two decimal places to get $129.07$ .

Steps

Step 1 ：Let's denote the degrees of freedom as $v$ and the z-score as $z_{0.975}$ . Given that $v = 100$ and $z_{0.975} = 1.959963984540054$ .

Step 2 ：Substitute these values into Fisher's approximation formula: $χ_{0.975}^{2} = \frac{(z_{0.975} + \sqrt{2 v - 1})^{2}}{2}$ .

Step 3 ：Calculate the value of $χ_{0.975}^{2}$ to get approximately 129.06942386990755.

Step 4 ：Round the result to two decimal places to get $129.07$ .