Question

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

x = # sold	probability
0	0.19
5	0.16
10	0.09
15	...?
20	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?
question help: message instructor post to forum

Explanation:

Step1: Recall probability sum rule

The sum of all probabilities in a probability - distribution is 1. Let $P(15)$ be the probability of selling 15 cheesecakes. Then $0.19 + 0.16+0.09 + P(15)+0.15 = 1$.

Step2: Solve for $P(15)$

$P(15)=1-(0.19 + 0.16+0.09 + 0.15)=1 - 0.59 = 0.41$.

Step3: Probability of selling at least 10 cheesecakes

$P(X\geq10)=P(10)+P(15)+P(20)=0.09 + 0.41+0.15 = 0.65$.

Step4: Probability of selling 5 or 15 cheesecakes

$P(X = 5\ or\ X = 15)=P(5)+P(15)=0.16 + 0.41 = 0.57$.

Step5: Probability of selling 25 cheesecakes

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

Step6: Calculate the expected value $E(X)$

The formula for the expected value of a discrete random variable is $E(X)=\sum_{i}x_ip_i$. So $E(X)=0\times0.19 + 5\times0.16+10\times0.09 + 15\times0.41+20\times0.15=0 + 0.8+0.9+6.15 + 3=10.85$.

Step7: Probability of selling at most 10 cheesecakes

$P(X\leq10)=P(0)+P(5)+P(10)=0.19 + 0.16+0.09 = 0.44$.

Answer:

• Probability of selling 15 cheesecakes: $0.41$
• Probability of selling at least 10 cheesecakes: $0.65$
• Probability of selling 5 or 15 cheesecakes: $0.57$
• Probability of selling 25 cheesecakes: $0$
• Expected number of cheesecakes sold in a day: $10.85$
• Probability of selling at most 10 cheesecakes: $0.44$