Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a random variable with mean 73 and standard deviation 10.5. How many students would have to take the examination to ensure, with probability at least 0.94, that the class average would be within 1.5 of 73?

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

[tex]X\sim N(\mu =73,\sigma =10.5)[/tex]

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

Solution to the problem

We want to find the value of n that satisfy this condition:

[tex]P(71.5 < \bar X <74.5)=0.94[/tex]

And we can use the z score formula given by:

[tex]z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And we have this:

[tex]P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z<\frac{74.5-73}{\frac{10.5}{\sqrt{n}}})[/tex]

And we can express this like this:

[tex]P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94[/tex]

And by properties of the normal distribution we can express this like this:

[tex]P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z<-0.14286 \sqrt{n})=0.94[/tex]

If we solve for [tex]P(Z<-0.14286 \sqrt{n})[/tex] we got:

[tex]P(Z<-0.14286 \sqrt{n})= 0.03[/tex]

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: [tex]z=-1.881[/tex]

And using this we have this equality:

[tex]-1.881 = -0.14286 \sqrt{n}[/tex]

If we solve for [tex]\sqrt{n}[/tex] we got:

[tex]\sqrt{n} = \frac{-1.881}{-0.14286}=13.167[/tex]

And then [tex]n=13.167^2 =173.369[/tex] and if we round up to the nearest integer we got n =174