Question

the cumulative gpa averages if all 6724 students at a crtain private university are normally distributed with a mean of 3.06 and standard deviation of 0.31. what is the probability that one student has a gpa of 3.75

Answer 1

The distribution is discrete with 6724 data points. It can be approximated with a normal cumulative probability distribution.
Using a cumulative distribution function the 'point-probability' that X is exactly b can be found as
\[P(X=b)=F(b)-\lim_{x \rightarrow b ^{-}}F(x)\]
With a normal distribution 95% of the data points lie within plus and minus 1.96 standard deviations from the mean. In this case 95% of the 6724 data points = 6388 data points. The interval for each data point as a decimal fraction of a standard deviation is given by
\[\frac{2\times 1.96}{6388}=0.0006\ standard\ deviation\]
The z-score for a gpa of 3.75 is
\[\frac{X-\mu}{\sigma}=\frac{3.75-3.06}{0.31}=2.2258\]
The z-score for the gpa in the adjacent interval below the gpa of 3.75 can be found from
\[2.2258-(1.5\times 0.0006)=2.2249\]
Using a standard normal distribution table the cumulative probabilities can be found as follows:
For z = 2.2258
\[p _{1}\approx 0.9870\]
For z = 2.2249
\[p _{2}\approx0.9869\]
The probability that one student has a gpa of 3.75 is
\[p _{1}-p _{2}\approx0.9870-0.9869\approx0.0001\]