Question

Anyone knows how to solve it?

What is the mean, variance and standard deviation of the sum of a 4 sided die and an 8 sided die? Use these values to estimate the mean, variance and standard deviation of the sum of 30, 4 sided dice plus 30, 8 sided dice. Plot the probability distribution of the estimated sum. What is the approximate probability of the sum being greater than 150? You may want to use graph paper for this question (or plot them with a computer).

Answer 1

`"The possible outcomes of throwing the 4 sided die are :"`
`"1, 2, 3, or 4. So the mean is (1+2+3+4)/4 = 2.5."`
`"The variance is equal to E[x²]-(E[x])² = (1²+2²+3²+4²)/4 -2.5²"`
`"= 30/4 - 2.5² = 7.5 - 6.25 = 1.25"`
`"The possible outcomes of throwing the 8 sided die are :"`
`"1, 2, 3, 4, 5, 6, 7, or 8. So the mean is 4.5."`
`"The variance is equal to (1²+2²+...+8²)/8 - 4.5² = 5.25."`
`"The mean of the sum of the two dice is the sum of the means,"`
`"so we have 2.5+4.5 = 7. "`
`"The variance is also the sum of the two variances :"`
`"1.25 + 5.25 = 6.5"`
`"The standard deviation is just the square root of the variance :"`
`"standard deviation = "sqrt(6.5)`

`"So if we have 30 4-sided dice and 30 8-sided dice, we get :"`
`"mean = 7*30 = 210"`
`"variance = 6.5 * 30 = 195"`
`"standard deviation = "sqrt(195)" = 13.964"`

`"The estimated sum will be approximately normally distributed"`
`"with mean 210 and standard deviation 13.964 :"`
`"N(210, 13.964)."`

`"P[sum > 150] ?"`
`"we go to the normalized normal distribution :"`
`"z = (149.5 - 210)/13.964 = -4.3325"`
`"(149.5 instead of 150 due to continuity correction)"`
`"we search this z value in a table for z values and find"`
`"a very small value, most tables plot only till -3.4 even."`
`"So P[sum > 150 ] = 0.9999..."`