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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.

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  • $\begingroup$ 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.) $\endgroup$ Mar 19 '19 at 6:05
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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 $P(X \leq 3)$: $$F(3)=\sum_{z=0}^3 {10 \choose 3} (0.4)^z (0.6)^{10-z}.$$

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  • $\begingroup$ I believe that it should be: $F(3)=\sum_{z=0}^3 {10 \choose 3} (0.4)^z (0.6)^{10-z}$. $\endgroup$ Mar 19 '19 at 1:44
  • $\begingroup$ Yes, thank you. $\endgroup$
    – dlnB
    Mar 19 '19 at 1:44

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