The new population of N = 5 scores with a mean of µ = 100 and a standard deviation of σ = 20 are 125, 105, 95, 90 and 135
(a) Compute µ and σ for the population.
The dataset is given as:
0, 6, 4, 3, and 12.
The mean is calculated as:
[tex]\mu = \frac{\sum x}n[/tex]
So, we have:
[tex]\mu = \frac{0 + 6 + 4 + 3 + 12}5[/tex]
[tex]\mu = 5[/tex]
The standard deviation is calculated as:
[tex]\sigma = \sqrt{\frac{\sum(x - \bar x)^2}n}[/tex]
This gives
[tex]\sigma = \sqrt{\frac{(0 - 5)^2 + (6- 5)^2 + (4- 5)^2 + (3- 5)^2 + (12- 5)^2}5[/tex]
[tex]\sigma = \sqrt{\frac{80}5[/tex]
[tex]\sigma = \sqrt{16[/tex]
[tex]\sigma = 4[/tex]
Hence, the values of μ and σ are μ = 5 and σ = 4
(b) The z-scores
This is calculated as:
[tex]z = \frac{x - \mu}{\sigma}[/tex]
When x = 0, 6, 4, 3, and 12.
We have:
[tex]z = \frac{0 - 5}{4} = 1.25[/tex]
[tex]z = \frac{6 - 5}{4} = 0.25[/tex]
[tex]z = \frac{4 - 5}{4} = -0.25[/tex]
[tex]z = \frac{3 - 5}{4} = -0.5[/tex]
[tex]z = \frac{12 - 5}{4} = 1.75[/tex]
Hence, the z-scores are 1.25, 0.25, -0.25, -0.5 and 1.75
(c) Transform the new population
We have:
N = 5, µ = 100 and σ = 20.
In (b), we have:
[tex]z = \frac{x - \mu}{\sigma}[/tex]
Make x the subject
[tex]x = \mu + z\sigma[/tex]
This gives
[tex]x_i = \mu + z_i\sigma[/tex]
So, we have:
[tex]x_1 = 100 + 1.25* 20 = 125[/tex]
[tex]x_2 = 100 + 0.25* 20 = 105[/tex]
[tex]x_3 = 100 - 0.25* 20 = 95[/tex]
[tex]x_4 = 100 - 0.5* 20 = 90[/tex]
[tex]x_5 = 100 + 1.75* 20 = 135[/tex]
Hence, the new population of N = 5 scores with a mean of µ = 100 and a standard deviation of σ = 20 are 125, 105, 95, 90 and 135
Read more about z-scores at:
https://brainly.com/question/16313918
