To calculate the expected value of the number of books ordered per student, we need to use the formula for the expected value in a discrete probability distribution:

$E(X) = \sum_{i=1}^{n} (x_i \cdot P(x_i))$

Where:

• $x_i$ is the value of the random variable (the number of books ordered),
• $P(x_i)$ is the probability of each outcome (the proportion of students who ordered $x_i$ books).

Step 1: Construct the frequency table

Number of Books Ordered ($x_i$)	Frequency ($f_i$)	Probability ($P(x_i)$)	$x_i \cdot P(x_i)$
0	18	$\frac{18}{200} = 0.09$	$0 \cdot 0.09 = 0$
1	27	$\frac{27}{200} = 0.135$	$1 \cdot 0.135 = 0.135$
2	64	$\frac{64}{200} = 0.32$	$2 \cdot 0.32 = 0.64$
3	59	$\frac{59}{200} = 0.295$	$3 \cdot 0.295 = 0.885$
4	32	$\frac{32}{200} = 0.16$	$4 \cdot 0.16 = 0.64$

Step 2: Calculate the expected value

Now, we compute the sum of $x_i \cdot P(x_i)$:

$E(X) = 0 + 0.135 + 0.64 + 0.885 + 0.64 = 2.3$

Thus, the expected value of the number of books ordered per student is 2.3.

Final Answer:

The expected number of books ordered per student is 2.3.

Would you like further clarification or have any other questions on this?

Here are some related questions:

1. How do you calculate the variance of the number of books ordered per student?
2. What is the standard deviation of the number of books ordered?
3. How would the expected value change if one more student ordered 4 books?
4. Can you use this method to find the expected value in a different type of problem?
5. How do you interpret the expected value in the context of this problem?

Tip: The expected value is a weighted average of all possible outcomes, where each outcome is weighted by its probability.