To solve this problem, we'll first need to understand the basic concepts of mean, variance, and standard deviation for a discrete random variable. Let's break it down step-by-step for clarity.
1. Mean (Expected Value):
The mean of a discrete random variable [tex]\(X\)[/tex] is found by multiplying each possible value of the variable by its probability and then summing all these products. Given the grades and their corresponding probabilities:
[tex]\[
\text{Mean} (\mu) = \sum (x_i \times P(x_i))
\][/tex]
Where [tex]\(x_i\)[/tex] are the grades and [tex]\(P(x_i)\)[/tex] are their respective probabilities.
2. Variance:
Variance measures the spread of the random variable's possible values around the mean. It is calculated using the formula:
[tex]\[
\sigma^2 = \sum ((x_i - \mu)^2 \times P(x_i))
\][/tex]
Where [tex]\(\mu\)[/tex] is the mean computed earlier.
3. Standard Deviation:
The standard deviation is simply the square root of the variance:
[tex]\[
\sigma = \sqrt{\sigma^2}
\][/tex]
Given the grades and probabilities as:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Grade} & 4 & 3 & 2 & 1 & 0 \\
\hline
\text{Probability} & 0.4 & 0.32 & 0.17 & 0.08 & 0.03 \\
\hline
\end{array}
\][/tex]
1. Calculate the Mean:
The mean grade ([tex]\(\mu\)[/tex]) is:
[tex]\[
(4 \times 0.4) + (3 \times 0.32) + (2 \times 0.17) + (1 \times 0.08) + (0 \times 0.03)
\][/tex]
This simplifies to:
[tex]\[
1.6 + 0.96 + 0.34 + 0.08 + 0 = 2.98
\][/tex]
2. Calculate the Variance:
The variance ([tex]\(\sigma^2\)[/tex]) is:
[tex]\[
((4 - 2.98)^2 \times 0.4) + ((3 - 2.98)^2 \times 0.32) + ((2 - 2.98)^2 \times 0.17) + ((1 - 2.98)^2 \times 0.08) + ((0 - 2.98)^2 \times 0.03)
\][/tex]
Simplifying the calculations results in:
[tex]\[
1.1596
\][/tex]
3. Calculate the Standard Deviation:
Taking the square root of the variance:
[tex]\[
\sigma = \sqrt{1.1596} = 1.076847
\][/tex]
Now, interpreting the result:
The correct interpretation of the standard deviation in the context of this problem is:
"The grade for a randomly selected student would typically vary from the expected grade by approximately 1.08."
Hence, the correct statement is:
"The grade for a randomly selected student would typically vary from the expected grade by 1.08."
