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If [tex]$X \sim N(25,31)$[/tex], find the [tex]$z$[/tex]-score for the data value 87.

A. [tex]$z=\frac{25-87}{31}=-2$[/tex]
B. [tex][tex]$z=\frac{31-87}{25}=-2.24$[/tex][/tex]
C. [tex]$z=\frac{87-31}{25}=2.24$[/tex]
D. [tex]$z=\frac{87-25}{31}=2$[/tex]


Sagot :

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].
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