Answered

The grade distribution for students in the introductory statistics class at a local community college is displayed in the table. In this table, [tex]$A=4, B=3$[/tex], etc. Let [tex]$X$[/tex] represent the grade for a randomly selected student.

\begin{tabular}{|c|c|c|c|c|c|}
\hline Grade & 4 & 3 & 2 & 1 & 0 \\
\hline Probability & 0.43 & [tex]$?$[/tex] & 0.17 & 0.05 & 0.04 \\
\hline
\end{tabular}

What is the probability that a randomly selected student got a B?

A. 0.17
B. 0.31
C. 0.43
D. 0.74



Answer :

GinnyAnswer
To determine the probability that a randomly selected student received a B, we'll follow these steps:

1. Identify the Known Probabilities:
The probabilities provided in the table are:
- Grade A (4): [tex]\( \text{P}(A) = 0.43 \)[/tex]
- Grade C (2): [tex]\( \text{P}(C) = 0.17 \)[/tex]
- Grade D (1): [tex]\( \text{P}(D) = 0.05 \)[/tex]
- Grade F (0): [tex]\( \text{P}(F) = 0.04 \)[/tex]

2. Understand the Total Probability Requirement:
The total probability of all possible outcomes must sum up to 1. That is:
[tex]\[ \text{P}(A) + \text{P}(B) + \text{P}(C) + \text{P}(D) + \text{P}(F) = 1 \][/tex]

3. Form the Equation:
We do not yet know [tex]\( \text{P}(B) \)[/tex]. We can represent it as [tex]\( x \)[/tex]. Plugging in the known values, we get:
[tex]\[ 0.43 + x + 0.17 + 0.05 + 0.04 = 1 \][/tex]

4. Simplify the Equation:
First, combine the known probabilities:
[tex]\[ 0.43 + 0.17 + 0.05 + 0.04 = 0.69 \][/tex]
So, the equation reduces to:
[tex]\[ 0.69 + x = 1 \][/tex]

5. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 1 - 0.69 = 0.31 \][/tex]

6. Conclusion:
The probability that a randomly selected student got a B is [tex]\( \text{P}(B) = 0.31 \)[/tex].

Hence, the correct option is:
[tex]\[ \boxed{0.31} \][/tex]

Other Questions