Question

a teacher asks her students to write down the number of hours studied, rounded to the nearest half hour. she compiles the results and develops the probability distribution below for a randomly selected student. what is the mean of the probability distribution?

hours studied (x): 0.5, 1, 1.5, 2, 2.5
probability (p(x)): 0.07, 0.2, 0.46, 0.2, 0.07

multiple choice options: 0.20, 0.46, 0.85, 1.50

Explanation:

Step1: Recall the formula for the mean of a discrete probability distribution

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

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

From the table:

• When \( x_1 = 0.5 \), \( P(x_1)=0.07 \)
• When \( x_2 = 1 \), \( P(x_2)=0.2 \)
• When \( x_3 = 1.5 \), \( P(x_3)=0.46 \)
• When \( x_4 = 2 \), \( P(x_4)=0.2 \)
• When \( x_5 = 2.5 \), \( P(x_5)=0.07 \)

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

• For \( x_1 = 0.5 \) and \( P(x_1)=0.07 \): \( 0.5\times0.07 = 0.035 \)
• For \( x_2 = 1 \) and \( P(x_2)=0.2 \): \( 1\times0.2 = 0.2 \)
• For \( x_3 = 1.5 \) and \( P(x_3)=0.46 \): \( 1.5\times0.46 = 0.69 \)
• For \( x_4 = 2 \) and \( P(x_4)=0.2 \): \( 2\times0.2 = 0.4 \)
• For \( x_5 = 2.5 \) and \( P(x_5)=0.07 \): \( 2.5\times0.07 = 0.175 \)

Step4: Sum up all the terms

\( \mu=0.035 + 0.2+0.69 + 0.4+0.175 \)
\( \mu=(0.035+0.2)+(0.69 + 0.4)+0.175 \)
\( \mu = 0.235+1.09+0.175 \)
\( \mu=0.235 + 1.265 \)
\( \mu = 1.5 \) (or \( 1.50 \) when rounded to two decimal places)

Answer:

1.50