Certainly! Let's carefully walk through the steps to find the [tex]\(z\)[/tex]-score for the data value 87, given the distribution [tex]\(X \sim N(25, 31)\)[/tex].
1. Identify the Given Values:
- Mean (µ) of the distribution: 25
- Standard deviation (σ) of the distribution: 31
- Data value (X): 87
2. Understand the Formula for [tex]\(z\)[/tex]-score:
The [tex]\(z\)[/tex]-score measures how many standard deviations a data point is away from the mean. The formula to calculate the [tex]\(z\)[/tex]-score is:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
where:
- [tex]\(X\)[/tex] is the data value
- [tex]\(\mu\)[/tex] is the mean
- [tex]\(\sigma\)[/tex] is the standard deviation
3. Substitute the Given Values into the Formula:
Let's substitute the given values (87 for [tex]\(X\)[/tex], 25 for [tex]\(\mu\)[/tex], and 31 for [tex]\(\sigma\)[/tex]) into the [tex]\(z\)[/tex]-score formula:
[tex]\[
z = \frac{87 - 25}{31}
\][/tex]
4. Calculate the [tex]\(z\)[/tex]-score:
- Subtract the mean from the data value:
[tex]\[
87 - 25 = 62
\][/tex]
- Divide the result by the standard deviation:
[tex]\[
z = \frac{62}{31}
\][/tex]
- When we perform the division, we get:
[tex]\[
z = 2
\][/tex]
Therefore, the [tex]\(z\)[/tex]-score for the data value 87, given the distribution [tex]\(X \sim N(25, 31)\)[/tex], is [tex]\(2\)[/tex].
