Each contestant in the Hunger Games must be trained to compete. Suppose that the time it takes to train a contestant has mean 10 days and standard deviation 2 days, independent of the time it takes other contestants to train. If the Hunger Games has 12 contestants to train, use the Central Limit Theorem to approximate the probability that it will take more than 150 days to train all the contestants. Leave your answer in terms of the standard normal distribution phi(a).

Suppose that the time it takes to train a contestant has mean 10 days and standard deviation 2 days, independent of the time it takes other contestants to train. Therefore,

[tex]E(X_i) = 10, Var(X_i) = 4[/tex]

Let [tex]X[/tex] be a random variable that counts the number of days needed to train all candidates. Thus,

[tex]X = \sum_{i=1}^{12} X_{i}[/tex]

Then, by taking expectation of both sides, we obtain

[tex]E(X) = E \left(\sum_{i=1}^{12} X_{i} \right)[/tex]

Because of the linearity of expectation, we obtain

[tex]E(X) = \sum_{i=1}^{12} E(X_{i})[/tex]

It is given that for all [tex]1 \leq i \leq 12[/tex]

[tex]E(X_i) = 10[/tex]

Now, we have that

[tex]E(X) = \sum_{i=1}^{12} E(X_{i}) = \sum_{i=1}^{12} 10 = 12 \cdot 10 = 120[/tex]

which means that the total time needed to train all candidates is 120 days.

Let's calculate the variance of [tex]X[/tex]. It is given that the time needed to train one candidate is independent of the time it takes other contestants to train. Hence, we know that [tex]X_i, 1 \leq i\leq 12[/tex] are independent. Therefore, the variance of a sum of 12 random variables equals the sum of variances of each of them and we obtain

[tex]Var(X) = Var\left( \sum_{i=1}^{12} X_i \right) = \sum_{i=1}^{12} Var(X_i)[/tex]

It is given that for all [tex]1 \leq i \leq 12[/tex]

[tex]Var(X_i) = 4[/tex]

Now, we have that

[tex]Var(X) = \sum_{i=1}^{12}Var(X_{i}) = \sum_{i=1}^{12} 4 = 12 \cdot 4 = 48[/tex]

Apply the Central Limit Theorem.

[tex]\dfrac{X- E(X)}{\sqrt{Var(X)} } = \dfrac{X- 120}{\sqrt{46} } : N(0,1)[/tex]

Now, let's calculate the probability that it will take more than 150 days to train all the contestants.

[tex]P(X> 150 ) = 1 - P(X\leq 150) \\\\\phantom{P(X> 150 ) }= 1- P\left( \dfrac{X- 120}{\sqrt{46} } \leq \dfrac{150- 120}{\sqrt{46} } \right)\\\\\phantom{P(X> 150 )} = 1- P\left( \dfrac{X- 120}{\sqrt{46} } \leq \dfrac{30}{\sqrt{46} } \right)\\\\\phantom{P(X> 150 )} = 1 - P(Z \leq 4.3301), Z:N(0,1)\\\\\phantom{P(X> 150 )} = 1 - \Phi(4.3301) \\\\\phantom{P(X> 150 )} = \Phi(-4.3301)[/tex]

Therefore, the approximation by the Central Limit Theorem that it will take more than 150 days to train all the contestants is

[tex]P(X>150) = \Phi(-4.3301)[/tex]