The constant 5 is one of a set of values satisfying the random variable [tex]$X \sim N(6,4)[tex]$[/tex]. What is the [tex]$[/tex]z$[/tex]-score for the value 5, correct to 2 decimal places?

Answer: [tex]\square[/tex]



Answer :

To determine the z-score for a given value in a normal distribution, you can use the formula for the z-score:

[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]

where:
- \( X \) is the value for which you want to find the z-score,
- \( \mu \) is the mean of the distribution,
- \( \sigma \) is the standard deviation of the distribution.

Given the random variable \( X \sim N(6,4) \), we know that the mean \( \mu = 6 \) and the variance \( \sigma^2 = 4 \). The standard deviation \( \sigma \) is the square root of the variance:

[tex]\[ \sigma = \sqrt{4} = 2 \][/tex]

Now, we need to find the z-score for the value \( X = 5 \):

1. Substitute the given values into the z-score formula:

[tex]\[ z = \frac{5 - 6}{2} \][/tex]

2. Calculate the numerator:

[tex]\[ 5 - 6 = -1 \][/tex]

3. Divide by the standard deviation:

[tex]\[ z = \frac{-1}{2} = -0.5 \][/tex]

Therefore, the z-score for the value 5 is \( -0.5 \) when rounded to 2 decimal places.

Answer: [tex]\(-0.5\)[/tex]