Question

the final exam grade distribution for all students in the introductory statistics class at a local community college is displayed in the table, with a = 4, b = 3, c = 2, d = 1, and f = 0. let x represent the grade for a randomly selected student from the class.
grade | 4 | 3 | 2 | 1 | 0
probability | 0.4 | 0.32 | 0.17 | 0.08 | 0.03
what is the mean of the distribution?
options: 3.50, 3, 2.98, 2

Explanation:

Step1: Recall the formula for the mean of a discrete random variable

The mean (expected value) \( \mu \) of a discrete random variable \( X \) is calculated as \( \mu=\sum_{i} x_{i}P(x_{i}) \), where \( x_{i} \) are the possible values of \( X \) and \( P(x_{i}) \) are their corresponding probabilities.

Step2: Identify the values of \( x_i \) and \( P(x_i) \)

From the table:

• When \( x = 4 \), \( P(4)=0.4 \)
• When \( x = 3 \), \( P(3)=0.32 \)
• When \( x = 2 \), \( P(2)=0.17 \)
• When \( x = 1 \), \( P(1)=0.08 \)
• When \( x = 0 \), \( P(0)=0.03 \)

Step3: Calculate each term \( x_iP(x_i) \)

• For \( x = 4 \): \( 4\times0.4 = 1.6 \)
• For \( x = 3 \): \( 3\times0.32 = 0.96 \)
• For \( x = 2 \): \( 2\times0.17 = 0.34 \)
• For \( x = 1 \): \( 1\times0.08 = 0.08 \)
• For \( x = 0 \): \( 0\times0.03 = 0 \)

Step4: Sum up all the terms

\( \mu=1.6 + 0.96+0.34 + 0.08+0 \)
First, \( 1.6+0.96 = 2.56 \)
Then, \( 2.56+0.34 = 2.9 \)
Then, \( 2.9+0.08 = 2.98 \)
Then, \( 2.98 + 0=2.98 \)

Answer:

\( 2.98 \)