Suppose 100 guests are playing a game at a party. Each guest rolls a dice once. If the dice roll result is 1, 3 or 5, that guest gets 1 gift. If the dice roll result is 2, 4 or 6, that guest gets 2 gifts. After the game, all the 100 guests put their gifts together to count the total number. Suppose the dice roll results are independent. Use Central Limit Theorem to approximate the probability that the total number of gifts is at least 160.

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Suppose 100 guests are playing a game at a party. Each guest rolls a dice once. If the dice
roll result is 1, 3 or 5, that guest gets 1 gift. If the dice roll result is 2, 4 or 6, that guest gets
2 gifts. After the game, all the 100 guests put their gifts together to count the total number.
Suppose the dice roll results are independent. Use Central Limit Theorem to approximate
the probability that the total number of gifts is at least 160.
Fact that may be useful: A dice has 6 sides: 1,2,3,4,5,6. When rolling a dice, each side
has an equal probability (1/6) of facing upwards. The number facing up is the dice roll
result. Your result can contain the CDF of N(0,1), Þ(·).
Transcribed Image Text:Suppose 100 guests are playing a game at a party. Each guest rolls a dice once. If the dice roll result is 1, 3 or 5, that guest gets 1 gift. If the dice roll result is 2, 4 or 6, that guest gets 2 gifts. After the game, all the 100 guests put their gifts together to count the total number. Suppose the dice roll results are independent. Use Central Limit Theorem to approximate the probability that the total number of gifts is at least 160. Fact that may be useful: A dice has 6 sides: 1,2,3,4,5,6. When rolling a dice, each side has an equal probability (1/6) of facing upwards. The number facing up is the dice roll result. Your result can contain the CDF of N(0,1), Þ(·).
Expert Solution
Step 1

From 100 guests are playing a game at a party.

Each guest rolls a dice once.

If the dice roll result is 1, 3 or 5, that guest gets 1 gift.

If the dice roll result is 2, 4 or 6, that guest gets 2 gifts.

P(1 gift)=P(1)+P(3)+P(5)

           =(1/6)+(1/6)+(1/6)

          =3/6

         =1/2

 

P(2 gifts)=P(2)+P(4)+P(6)

           =(1/6)+(1/6)+(1/6)

          =3/6

         =1/2

Let X denote the number of gifts and PMF of X is given below:

X 1 2
P(X) 0.5 0.5

 

Consider,

μ=XPX  =1*0.5+2*0.5 =0.5+1 =1.5

EX2=X2PX  =12*0.5+22*0.5 =0.5+2 =2.5

σ2=EX2-μ2    =2.5-1.5    =1

 

 

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