1 A candy store consists of a 1 dollar candy bar 5 dollar ca
1) A candy store consists of a 1 dollar candy bar, 5 dollar candy bar and a 9 dollar candy bar. Construct the sampling distribution for the mean price of candy bars using sample size 2 taken with replacement.
2) A meeting involves three women Owners , six men Owners, five women Presidents, four men Presidents, eight women Assistants and five men Assistants. If one occupant is selected random from the meeting, find the probability that the occupant is a Man or a President.
Solution
(1)
Let A,B,C respectively denote that $1,$5 , $9 candy is selected.
Assume that P(A)=P(B)=P(C)=1/3 and they are independent.
price(A)=1, price(B)=5, price(C)=9
In case of with replacement sample of size 2, there may be 32= 9 possible cases viz.
AA,AB,BA,BB,AC,CA,CC,BC,CB each with probability 1/9.
Let X denote the mean price.
Then we construct the following table:
Hence we see that X can take the values 1,3,5,7,9. The sampling distribution of X is
(2)
Required to find
P(the occupant is a man or a president)
=P(the occupant is a man) + P(the occupant is a president) - (the occupant is man president)
=(6+4+5) / (3+6+5+4+8+5) + (5+4) / (3+6+5+4+8+5) - 4 / (3+6+5+4+8+5)
=15/31 + 9/31 - 4/31
=20/31
| Outcome | Value of X, average price | Probability |
| AA | (1+1)/2 = 1 | 1/9 |
| AB | (1+5)/2 = 3 | 1/9 |
| BA | (5+1)/2=3 | 1/9 |
| BB | (5+5)/2 =5 | 1/9 |
| AC | (1+9)/2 = 5 | 1/9 |
| CA | (9+1)/2 = 5 | 1/9 |
| CC | (9+9)/2 =9 | 1/9 |
| BC | (5+9)/2 =7 | 1/9 |
| CB | (9+5)/2 = 7 | 1/9 |
