A die is rolled 24 times. Use the Central Limit Theorem to estimate the probability that

  1. the sum is greater than 84.

\[μ = E(X)=1(\frac{1}{6}) +2 (\frac{1}{6}) +3(\frac{1}{6})+4 (\frac{1}{6}) +5 (\frac{1}{6}) +6 (\frac{1}{6}) = (\frac{7}{2})\] To find the variance of X, we form the new random variable (X − μ)2 and compute its expectation.

\[V(X)= \frac{1}{6}(\frac{25}{4} + \frac{9}{4} + \frac{1}{4} + \frac{1}{4} +\frac{9}{4} + \frac{25}{4}) = \frac{35}{12}\]

\[P(S_{24} > 84) = P(S_{24} > \frac{84.5 - 84}{8.366})\]

Ex = 7/2
Vx = 35/12
ES24 = 24 * Ex
VS24 = 24 * Vx
sqrtVS24 = sqrt(VS24)
P = (84.5 - ES24)/sqrtVS24
Percentile = 1 - pnorm(P)

paste("P(S24>84) = ", Percentile)
## [1] "P(S24>84) =  0.47617282233683"
  1. the sum is equal to 84.

Using the same variables calculated above…

\[P(S_{24} > 84) = P( \frac{83.5 - 84}{8.366} > S_{24} > \frac{84.5 - 84}{8.366})\]

Phigh = (84.5 - ES24)/sqrtVS24
Plow = (83.5 - ES24)/sqrtVS24
Percent_high = pnorm(Phigh)
Percent_low = pnorm(Plow)

paste("P(S24>84) = ", Percent_high - Percent_low)
## [1] "P(S24>84) =  0.0476543553263407"