A population consists of the following N = 5 scores

A population consists of the following N = 5 scores: 0, 6, 4, 3, and 12.

a. Compute µ and u for the population.

b. Find the z-score for each score in the population.

c. Transform the original population into a new population of N = 5 scores with a mean of

µ= 100 and a standard deviation of u= 20.

The correct answer and explanation is :

Let’s go step by step to compute and explain each part of the problem.

a. Compute µ (Mean) and σ (Standard Deviation) for the population

To find the mean (µ) of the population, we use the formula:

[
\mu = \frac{1}{N} \sum_{i=1}^{N} x_i
]

Where:

• ( N = 5 ) (the number of scores),
• ( x_1 = 0, x_2 = 6, x_3 = 4, x_4 = 3, x_5 = 12 ) (the scores).

So, we compute the sum of the scores first:

[
\sum_{i=1}^{5} x_i = 0 + 6 + 4 + 3 + 12 = 25
]

Now, divide the sum by ( N ) to get the mean:

[
\mu = \frac{25}{5} = 5
]

Now, for the standard deviation (σ), we use the formula:

[
\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i – \mu)^2}
]

We first compute the squared deviations from the mean:

[
(0 – 5)^2 = 25, \quad (6 – 5)^2 = 1, \quad (4 – 5)^2 = 1, \quad (3 – 5)^2 = 4, \quad (12 – 5)^2 = 49
]

Sum of squared deviations:

[
25 + 1 + 1 + 4 + 49 = 80
]

Now, divide the sum by ( N ) and take the square root:

[
\sigma = \sqrt{\frac{80}{5}} = \sqrt{16} = 4
]

So, the mean (µ) is 5, and the standard deviation (σ) is 4.

b. Find the z-score for each score in the population

The z-score for each score in the population is calculated using the formula:

[
z = \frac{x_i – \mu}{\sigma}
]

For each score:

1. For ( x_1 = 0 ):
   [
   z = \frac{0 – 5}{4} = \frac{-5}{4} = -1.25
   ]
2. For ( x_2 = 6 ):
   [
   z = \frac{6 – 5}{4} = \frac{1}{4} = 0.25
   ]
3. For ( x_3 = 4 ):
   [
   z = \frac{4 – 5}{4} = \frac{-1}{4} = -0.25
   ]
4. For ( x_4 = 3 ):
   [
   z = \frac{3 – 5}{4} = \frac{-2}{4} = -0.5
   ]
5. For ( x_5 = 12 ):
   [
   z = \frac{12 – 5}{4} = \frac{7}{4} = 1.75
   ]

Thus, the z-scores for the population are:

• For 0: ( z = -1.25 )
• For 6: ( z = 0.25 )
• For 4: ( z = -0.25 )
• For 3: ( z = -0.5 )
• For 12: ( z = 1.75 )

c. Transform the original population into a new population with a mean of 100 and a standard deviation of 20

To transform the scores to a new population with a specified mean ( \mu = 100 ) and standard deviation ( \sigma = 20 ), we use the formula for transformation:

[
x_{\text{new}} = \mu_{\text{new}} + z \cdot \sigma_{\text{new}}
]

Where:

• ( \mu_{\text{new}} = 100 ),
• ( \sigma_{\text{new}} = 20 ),
• The z-scores are calculated in part (b).

Now, transform each score:

1. For ( x_1 = 0 ) (with ( z = -1.25 )):
   [
   x_{\text{new}} = 100 + (-1.25) \cdot 20 = 100 – 25 = 75
   ]
2. For ( x_2 = 6 ) (with ( z = 0.25 )):
   [
   x_{\text{new}} = 100 + 0.25 \cdot 20 = 100 + 5 = 105
   ]
3. For ( x_3 = 4 ) (with ( z = -0.25 )):
   [
   x_{\text{new}} = 100 + (-0.25) \cdot 20 = 100 – 5 = 95
   ]
4. For ( x_4 = 3 ) (with ( z = -0.5 )):
   [
   x_{\text{new}} = 100 + (-0.5) \cdot 20 = 100 – 10 = 90
   ]
5. For ( x_5 = 12 ) (with ( z = 1.75 )):
   [
   x_{\text{new}} = 100 + 1.75 \cdot 20 = 100 + 35 = 135
   ]

So, the new population of scores is: 75, 105, 95, 90, 135.

Explanation of Transformations

To standardize or transform a population, the z-score gives us a way to express how far each value is from the mean in terms of the standard deviation. In part (a), we calculated the mean and standard deviation of the original population, and in part (b), we calculated the z-scores for each score, which standardize them by removing the effect of the scale and center. In part (c), we used those z-scores to transform the original population into a new one, scaling the values to have a specified mean and standard deviation, thus creating a new distribution with the desired properties.