Hi Gladys,
Probability of selling 15 cheesecakes
You were given probabilities for all other cake counts except 15. In discrete probability distribution, all probabilities must add to 1, so, if we treat cakes sold as X:
P(X=15)=1-(0.1+0.21+0.04+0.13)=0.52
Probability of Selling 5 or 15
Again, if we treat as discrete--no values for X possible other than those given--note this is unrealistic, it's certainly possible to sell 6 cakes, 8 cakes, etc.
P(X=5 or X=15)=P(X=5)+P(X=15)=
0.21+0.52
P(X=5 or X=15)=0.73
Probability of Selling 25
We cannot tell from the information given. We have no value above 15, so we can't measure for 25. I'm curious about what the blank value is for X in the last column.
Expected Number of Cheesecakes Sold in a Day
Again, we need that last X-value in the last column. If I assume it is 20, we can proceed with this. The appropriate formula is:
E(X)=Sum(x(P(x))
where x=cakes sold--only the numbers you were given
P(x)=Probability that given number of cakes were sold. We were given:
P(0)=0.1
P(5)=0.21
P(10)=0.04
P(15)=0.52, from previous computation
P(20)=0.13, assumed X value in last column
E(X)=[0.1(0)] + [0.21(5)] + [0.04(10)] + [0.52(15)] + [0.13(20)]
E(X)=12.3 cakes
Probability of Selling at most 10
Again, we have three discrete probabilities here, P(X=0), P(X=5), and P(X=10). You were given all three, so:
P(X<=10)= P(X=0) + P(X=5) + P(X=10)
P(X<=10)=0.1+0.21+0.04
P(X<=10)= 0.35
I hope this helps.
