Question

a local bakery has determined a probability distribution for the number of cheesecakes that they sell in a given day.
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 at most 10 cheesecakes?
give the expected number of cheesecakes sold in a day using the discrete probability distribution.

Explanation:

Step1: Find the missing probability

Since the sum of all probabilities in a probability distribution is \(1\). Let the missing probability be \(p\). Then \(0.22 + 0.05+0.2 + p+0.11 = 1\).

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Step2: Probability of selling 15 cheesecakes

The probability of selling \(15\) cheesecakes is the value of \(p\) we just found, which is \(0.42\)

Step3: Probability of selling at most 10 cheesecakes

\(P(X\leq10)=P(X = 0)+P(X = 5)+P(X = 10)=0.22+0.05 + 0.2=0.47\)

Step4: Probability of selling at least 10 cheesecakes

\(P(X\geq10)=P(X = 10)+P(X = 15)+P(X = 20)=0.2+0.42+0.11 = 0.73\)

Step5: Expected value

The formula for the expected value \(E(X)=\sum_{i}x_ip_i\)

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Answer:

• Probability of selling \(15\) cheesecakes: \(0.42\)
• Probability of selling at most \(10\) cheesecakes: \(0.47\)
• Probability of selling at least \(10\) cheesecakes: \(0.73\)
• Expected number of cheesecakes sold: \(10.75\)