Calculating the z Score for 95% Confidence Level - Mathematical Method | Sirsh

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To find the z score for a 95% confidence level, one must determine the value of z such that the area under the standard normal distribution between -z and z equals 0.95. This involves using the standard normal distribution formula, which includes the integral of the probability density function. The calculated z score for this confidence level is approximately 1.9600. Understanding this requires knowledge of the properties of the normal distribution and its relation to confidence intervals. This mathematical approach confirms the z score's significance in statistical analysis.
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Hi, I'ld like to know if anyone knows how to find the 'z' score inrelation to a percentage degree of confidence. The question is: Determine the z score that would give a result with a degree of confidence of 95%.

I know that it is, 1.9600. However, I would like to know how to figure this out mathematically.

Thanks,
Sirsh.
 
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Sirsh said:
Hi, I'ld like to know if anyone knows how to find the 'z' score inrelation to a percentage degree of confidence. The question is: Determine the z score that would give a result with a degree of confidence of 95%.

I know that it is, 1.9600. However, I would like to know how to figure this out mathematically.

Thanks,
Sirsh.

It's based on the area under the standard normal distribution:
\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z^2}

Your question "Determine the z score that would give a result with a degree of confidence of 95%." comes down to "what is the value of z such that the area under the standard normal distribution between -z and z is equal to 0.95?". And it turns out that:

\int_{-1.96}^{1.96} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z^2} dz \approx 0.95
 
Thanks alot, gerben!
 
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