Question

Please answer these questions. and show detail.

1: Your new job is to test the fairness of dice in Las Vegas.  What is the probability (in decimal) that one will roll a 1 with a 6 sided die?

2: What is the probability that one will roll 3 1s in 6 attempts?

3: Given a normal distribution what is the probability one would get a value greater than 3 in a population with a mean of 1 and SD of 1?

4: Given the same conditions as in the question above what is the probability that one would get a value less than -1?

5: Given the same normal distribution what is the median value?

Answer 1

» If a 6-sided die is rolled, then the sample space is s={1,2,3,4,5,6} . .: The total number of elementary events is n = 6. Let that with I denotes the event one will roll a 1 a 6-sided die, inlA= 1 [ If E is an event, then ALE) is the number of elementary events

1 favourable to E. Here the only possible elementary event is : A = {13. N1A)=1. . Required probability - PLA) = n(A) = - - 0.17. n Let X denores the number of Is. Here total number of rolls is n=6. Let o denotes the probability of getting a P= I. We assume that the rolls are independent

X ~ Bin (n=6, P = 16 ) The pmf of x is PLX= x) = f( t) " ( ) 6-1 n=0(1) 6 0.W . : Required probability = P(x =3) 6-3 -04 () = 0.05358.

3) Let y follows normal distribution with a mean of 1 and a SD of 1. .. Y~ N(1,1) .. Required proobability = ply> 3) = P(Z > 2) [ her z=Y-1, then 2 ~ N(0,1). = I-Plz =2) = 1- Ø12) [ $(2) is the edf of N10,). - 0 9772 = 0.0 228.

4) Required probability, ) = PlyC-1) PLY = PLZ 2-2) - $(-2) = 1- $(2) [ If koo, then $(-k) = 1 - 0(k). 7 - 1- 0.9772 = 0.0228. 5) we know that if a random variable X~NIM, 52), then the median of x is Me=M. Here M=1. ["Y~N(1,1)] The median of y = Me = M = 1.