Question

a local bakery has determined a probability distribution for the number of cheesecakes that they sell in a given day.
x = #sold 0 5 10 15 20
probability 0.17 0.13 0.16 ...? 0.15
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?

Explanation:

Step1: Find missing probability

The sum of all probabilities in a probability - distribution is 1. Let the probability of selling 15 cheesecakes be $p$. Then $0.17 + 0.13+0.16 + p+0.15 = 1$. Solving for $p$:
\[p=1-(0.17 + 0.13+0.16 + 0.15)=1 - 0.61 = 0.39\]

Step2: Probability of selling at least 10 cheesecakes

At least 10 means 10, 15, or 20. So $P(X\geq10)=P(X = 10)+P(X = 15)+P(X = 20)=0.16 + 0.39+0.15=0.7$

Step3: Probability of selling 5 or 15 cheesecakes

Using the addition rule for mutually - exclusive events, $P(X = 5\ or\ X = 15)=P(X = 5)+P(X = 15)=0.13 + 0.39 = 0.52$

Step4: Probability of selling 25 cheesecakes

Since 25 is not in the given distribution, $P(X = 25)=0$

Step5: Calculate the expected value

The formula for the expected value $E(X)$ of a discrete random variable is $E(X)=\sum_{i}x_ip_i$. So $E(X)=0\times0.17+5\times0.13 + 10\times0.16+15\times0.39+20\times0.15=0 + 0.65+1.6+5.85 + 3=11.1$

Step6: Probability of selling at most 10 cheesecakes

At most 10 means 0, 5, or 10. So $P(X\leq10)=P(X = 0)+P(X = 5)+P(X = 10)=0.17+0.13 + 0.16=0.46$

Answer:

Probability of selling 15 cheesecakes: 0.39
Probability of selling at least 10 cheesecakes: 0.7
Probability of selling 5 or 15 cheesecakes: 0.52
Probability of selling 25 cheesecakes: 0
Expected number of cheesecakes sold in a day: 11.1
Probability of selling at most 10 cheesecakes: 0.46