Can you help me on some question i'm stuck on?
Let X be a random variable of the average grade of the students in some course.
E[X] = 75
Var[X] = 25
What's the number of students that need to be checked in order to assure in at least 90% that their average grade is between 70 and 80?
Here's what i've done:
$P(70<X<80) < P(|X-75|<5) = 1-P(|X-75|>=5) >= 0.9$ (according to Chebyshev)
Now let $5=a*s$ (s = standard deviation , a is a parameter). Then we need:
$1-\frac{1}{a^2} = 0.9$
That is $a=\sqrt{10}$
Which means $s=\frac{5}{\sqrt{10}}$
Which means Variance = $s^2 = 2.5$
So apparently this is wrong because i believe i can't change the variance in the question... Plus, i didn't conclude anything about the number of students...
Any help?
Thanks
We have a population mean of $\mu=75$ and variance of $\sigma^2=25$. So the distribution of the sample average form a sample of size $n$ also has mean $75$ and has a variance of $\frac{25}{n}$.
So if you decided to use Chebyshev's inequality of the form $\Pr(|Y-E[Y]| \ge t] \le \text{Var}(Y)/t^2$ you would want $ \dfrac{\frac{25}{n}}{5^2} \lt 0.1$, i.e. a sample size of $n \gt 10$.
If you assumed a normal distribution, you would want to look at $ \sqrt{\frac{25}{n}} \Phi^{-1}(0.95) \lt 5$ which would lead to a sample size $n \gt 2.706$.