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 0.25 0.16 0.01...? 0.1

a. what is the probability of selling 15 cheesecakes in a given day?

b. what is the probability of selling at least 10 cheesecakes?

c. what is the probability of selling 5 or 15 cheesecakes?

d. what is the probability of selling 25 cheesecakes?

e. give the expected number of cheesecakes sold in a day using the discrete probability distribution?

f. what is the probability of selling at most 10 cheesecakes?

Explanation:

Step1: Find probability for selling 15 cheesecakes

The sum of all probabilities in a probability distribution is 1. Let \(P(15)=x\). Then \(0.25 + 0.16+0.01+x + 0.1=1\).

$$x=1-(0.25 + 0.16+0.01 + 0.1)=1 - 0.52=0.48$$

Step2: Probability of selling at least 10 cheesecakes

At - least 10 means \(P(X = 10)+P(X = 15)+P(X = 20)\).

$$P(X\geq10)=0.01 + 0.48+0.1=0.59$$

Step3: Probability of selling 5 or 15 cheesecakes

Using the addition rule for mutually - exclusive events \(P(A\cup B)=P(A)+P(B)\). Here \(A\) is selling 5 cheesecakes and \(B\) is selling 15 cheesecakes.

$$P(X = 5\cup X = 15)=0.16+0.48 = 0.64$$

Step4: Probability of selling 25 cheesecakes

Since 25 is not in the range of \(X\) ( \(X=\{0,5,10,15,20\}\) ), \(P(X = 25)=0\)

Step5: Calculate the expected value

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

$$E(X)=(0\times0.25)+(5\times0.16)+(10\times0.01)+(15\times0.48)+(20\times0.1)$$

$$E(X)=0 + 0.8+0.1+7.2 + 2$$

$$E(X)=10.1$$

Step6: Probability of selling at most 10 cheesecakes

At - most 10 means \(P(X=0)+P(X = 5)+P(X = 10)\)

$$P(X\leq10)=0.25+0.16 + 0.01=0.42$$

Answer:

a. \(0.48\)
b. \(0.59\)
c. \(0.64\)
d. \(0\)
e. \(10.1\)
f. \(0.42\)