Questions: The number of cheesecakes sold each day by a bakery follows a random variable X. The distribution of X is in the table below: x 0 5 10 15 20 P(x) 0.11 0.11 0.18 (?) 0.07 On a given day, what is the probability that the bakery sells... A) .15 cheesecakes? 0.53 B) ... at least 10 cheesecakes? 0.78 C) ... at most 10 cheesecakes? 0.40 D) ... 5 or 15 cheesecakes? E) ... 25 cheesecakes? F) What is the expected value of the number of cheesecakes sold daily? 11.7

The number of cheesecakes sold each day by a bakery follows a random variable X. The distribution of X is in the table below: x 0 5 10 15 20 P(x) 0.11 0.11 0.18 (?) 0.07 On a given day, what is the probability that the bakery sells... A) .15 cheesecakes? 0.53 B) ... at least 10 cheesecakes? 0.78 C) ... at most 10 cheesecakes? 0.40 D) ... 5 or 15 cheesecakes? E) ... 25 cheesecakes? F) What is the expected value of the number of cheesecakes sold daily? 11.7
Transcript text: The number of cheesecakes sold each day by a bakery follows a random variable $X$. The distribution of $X$ is in the table below: \begin{tabular}{|c|c|c|c|c|c|} \hline$x$ & 0 & 5 & 10 & 15 & 20 \\ \hline$P(x)$ & 0.11 & 0.11 & 0.18 & $(?)$ & 0.07 \\ \hline \end{tabular} On a given day, what is the probability that the bakery sells... A). .15 cheesecakes? \[ 0.53 \] B) $\ldots$ at least 10 cheesecakes? \[ 0.78 \] C) $\ldots$ at most 10 cheesecakes? \[ 0.40 \] D) $\ldots 5$ or 15 cheesecakes? $\square$ E) ... 25 cheesecakes? $\square$ F) What is the expected value of the number of cheesecakes sold daily? \[ 11.7 \]
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Solution

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Solution Steps

Step 1: Update the Probability Distribution

The given probability distribution for the number of cheesecakes sold each day is as follows:

\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & 5 & 10 & 15 & 20 \\ \hline P(x) & 0.11 & 0.11 & 0.18 & ? & 0.07 \\ \hline \end{array} \]

To find the missing probability \( P(15) \), we use the fact that the sum of all probabilities must equal 1:

\[ P(15) = 1 - (0.11 + 0.11 + 0.18 + 0.07) = 1 - 0.47 = 0.53 \]

Thus, the updated probability distribution is:

\[ \begin{array}{|c|c|c|c|c|c|} \hline x & 0 & 5 & 10 & 15 & 20 \\ \hline P(x) & 0.11 & 0.11 & 0.18 & 0.53 & 0.07 \\ \hline \end{array} \]

Step 2: Calculate the Probability of Selling at Least 10 Cheesecakes

To find the probability of selling at least 10 cheesecakes, we sum the probabilities for \( x = 10, 15, \) and \( 20 \):

\[ P(X \geq 10) = P(10) + P(15) + P(20) = 0.18 + 0.53 + 0.07 = 0.78 \]

Step 3: Calculate the Probability of Selling at Most 10 Cheesecakes

To find the probability of selling at most 10 cheesecakes, we sum the probabilities for \( x = 0, 5, \) and \( 10 \):

\[ P(X \leq 10) = P(0) + P(5) + P(10) = 0.11 + 0.11 + 0.18 = 0.40 \]

Step 4: Calculate the Expected Value

The expected value \( E(X) \) of the number of cheesecakes sold daily is calculated as follows:

\[ E(X) = \sum (x \cdot P(x)) = 0 \times 0.11 + 5 \times 0.11 + 10 \times 0.18 + 15 \times 0.53 + 20 \times 0.07 \]

Calculating each term:

\[ E(X) = 0 + 0.55 + 1.8 + 7.95 + 1.4 = 11.7 \]

Final Answer

The answers to the questions are as follows:

  • A) The probability of selling at least 10 cheesecakes is \( 0.78 \).
  • B) The probability of selling at most 10 cheesecakes is \( 0.40 \).
  • C) The expected value of the number of cheesecakes sold daily is \( 11.7 \).

Thus, the final boxed answers are:

\[ \boxed{0.78} \] \[ \boxed{0.40} \] \[ \boxed{11.7} \]

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