A die is thrown until the first time the total sum of the face values of the die is 700 or greater. Estimate the probability that, for this to happen,
\(\mu = E(X) = \frac{(1+2+3+4+5+6)}{6} = 3.5\)
\(\sigma = \sqrt{\frac{(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2}{6}} = 1.708\)
\(\sigma ^2 = V(X) = 2.92\)
more than 210 tosses are required
\(E(X_{211}) = 3.5 \times 211 = 738.5\)
\(V(X_{211}) = 2.92 \times 211 = 616.12\)
\(\sigma = \sqrt{616.12} = 24.8\)
\(P(X_{211} > 700) = P(\frac{700-738.5}{24.8}) = -1.55\)
pnorm(-1.55)## [1] 0.06057076less than 190 tosses are required
\(E(X_{189}) = 3.5 \times 189 = 661.5\)
\(V(X_{189}) = 2.92 \times 189 = 551.88\)
\(\sigma = 23.5\)
\(P(X_{189} < 700) = P(\frac{700-661.5}{23.5}) = 1.64\)
1 - pnorm(1.64)## [1] 0.05050258between 180 and 210 tosses, inclusive, are required
\(E(X_{180}) = 3.5 \times 180 = 630\)
\(E(X_{210}) = 3.5 \times 210 = 735\)
\(V(X_{180}) = 2.92 \times 180 = 525.6\)
\(V(X_{210}) = 2.92 \times 210 = 613.2\)
\(\sigma _{180} = 22.93\)
\(\sigma _{210} = 24.76\)
\(P(X_{180} < 700) = P(\frac{700-630}{22.93}) = 3.05\)
\(P(X_{210} < 700) = P(\frac{700-735}{24.76}) = -1.41\)
# P(210) - P(180)
(1 - pnorm(-1.41)) - (1 - pnorm(3.05))## [1] 0.919586