# Is this a t-distribution problem?

The problem says that the probability of student failure is 0.4 A random sample of size 10 is taken from this population. I'm asked to find the probability that, at most, 30 % of students failed.

I don't understand how to start but I'll try:

sample size $$n = 10$$, then the degrees of freedom $$\nu = n -1 = 9$$, if the probability of one student failing is $$0.4$$, then the expected value of the sample $$\bar x = (0.4) (10) = 4$$?

I'm not sure about that last part, now to find the probability of at most 30 % students failed, should I calculate:

$$P( T <0.3)$$ ? I'm really lost, I only understand the difference between the t-distribution and the normal distribution, but don't understand how to solve these kinds of problems or how to make sense out of them, hopefully you can help me, thanks.

• A problem that follows a pass/fail pattern is rather like a problem that follows a heads/tails pattern. Which probability distribution did you use for "fewer than 3 tails" in a set of $n=10$ coin flips with $p(heads)=0.6$? (Hint: this has nothing to do with t-distributions.) Mar 19 '19 at 6:05

This is a binomial $$(10,.4)$$ distribution. If $$X \sim binomial(n,p)$$, then it's probability mass function is $$f(x)={n \choose x} p^x(1-p)^{n-x},$$ which for your parameters is $$f(x)={10 \choose x} (0.4)^x(0.6)^{10-x}.$$ The cdf is then $$F(x) = \sum_{z=0}^x f(z)=\sum_{z=0}^x {10 \choose x} (0.4)^z(0.6)^{10-z}.$$
30% of the sample is 3 individuals here. Hence we want $$P(X \leq 3)$$: $$F(3)=\sum_{z=0}^3 {10 \choose 3} (0.4)^z (0.6)^{10-z}.$$
• I believe that it should be: $F(3)=\sum_{z=0}^3 {10 \choose 3} (0.4)^z (0.6)^{10-z}$. Mar 19 '19 at 1:44