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]
[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]
