Question

a local bakery has determined a probability distribution for the number of cheesecakes that they sell in a given day.

x = # sold	probability
5	0.25
10	0.16
15	...?
20	0.18

what is the probability of selling 15 cheesecakes in a given day?
what is the probability of selling at least 10 cheesecakes?
what is the probability of selling 5 or 15 cheesecakes?
what is the probability of selling 25 cheesecakes?
give the expected number of cheesecakes sold in a day using the discrete probability distribution?
what is the probability of selling at most 10 cheesecakes?
question help: message instructor

Explanation:

Step1: Find probability of selling 15 cheesecakes

The sum of all probabilities in a probability - distribution is 1. Let \(P(15)\) be the probability of selling 15 cheesecakes. Then \(0.29 + 0.25+0.16 + P(15)+0.18 = 1\). So \(P(15)=1-(0.29 + 0.25+0.16 + 0.18)=1 - 0.88 = 0.12\).

Step2: Probability of selling at least 10 cheesecakes

At least 10 means 10, 15, or 20. So \(P(X\geq10)=P(10)+P(15)+P(20)=0.16 + 0.12+0.18 = 0.46\).

Step3: Probability of selling 5 or 15 cheesecakes

For mutually - exclusive events, \(P(A\cup B)=P(A)+P(B)\). So \(P(5\ or\ 15)=P(5)+P(15)=0.25 + 0.12 = 0.37\).

Step4: Probability of selling 25 cheesecakes

Since 25 is not in the given probability distribution, \(P(25)=0\).

Step5: Calculate the expected value

The expected value \(E(X)=\sum_{i}x_{i}P(x_{i})\), where \(x_{i}\) are the number of cheesecakes sold and \(P(x_{i})\) are their probabilities. \(E(X)=0\times0.29 + 5\times0.25+10\times0.16 + 15\times0.12+20\times0.18=0 + 1.25+1.6+1.8+3.6 = 8.25\).

Step6: Probability of selling at most 10 cheesecakes

At most 10 means 0, 5, or 10. So \(P(X\leq10)=P(0)+P(5)+P(10)=0.29 + 0.25+0.16 = 0.7\).

Answer:

Probability of selling 15 cheesecakes: 0.12
Probability of selling at least 10 cheesecakes: 0.46
Probability of selling 5 or 15 cheesecakes: 0.37
Probability of selling 25 cheesecakes: 0
Expected number of cheesecakes sold: 8.25
Probability of selling at most 10 cheesecakes: 0.7