This question involves understanding how linear transformations affect the mean and standard deviation of a dataset. We are given a set of observations, $x_i$, and a transformed set, $w_i$, related by the equation $w_i = p x_i + k$. We know the mean and standard deviation of $x_i$ and $w_i$, and we need to find the constants $p$ and $k$.

Understanding Linear Transformation Properties

For a linear transformation of the form $Y = aX + b$, where $a$ and $b$ are constants, the following properties hold for the mean ($E$) and standard deviation ($SD$):

• Mean Transformation: The mean of the transformed variable $Y$ is related to the mean of the original variable $X$ by the formula: $E[Y] = a E[X] + b$.
• Standard Deviation Transformation: The standard deviation of the transformed variable $Y$ is related to the standard deviation of the original variable $X$ by the formula: $SD[Y] = |a| SD[X]$. The absolute value is used because standard deviation cannot be negative.

Applying Properties to the Problem

In this problem, we have $w_i = p x_i + k$. Comparing this to $Y = aX + b$, we have $Y = w_i$, $X = x_i$, $a = p$, and $b = k$.

We are given:

• Mean of $x_i$, denoted as $E[x] = 48$.
• Standard deviation of $x_i$, denoted as $SD[x] = 12$.
• Mean of $w_i$, denoted as $E[w] = 55$.
• Standard deviation of $w_i$, denoted as $SD[w] = 15$.

Calculating the Constant 'p'

We use the standard deviation transformation property:

$$SD[w] = |p| SD[x]$$

Substitute the given values:

$$15 = |p| \times 12$$

Now, solve for $|p|$:

$$|p| = \frac{15}{12}$$

Simplify the fraction:

$$|p| = \frac{5}{4}$$

$$|p| = 1.25$$

This implies that $p$ could be either $1.25$ or $-1.25$. We need to use the mean transformation property to determine the correct value of $p$ and find $k$.

Calculating the Constant 'k'

We use the mean transformation property:

$$E[w] = p E[x] + k$$

Substitute the given values:

$$55 = p \times 48 + k$$

Rearrange the equation to solve for $k$:

$$k = 55 - 48 \times p$$

Now, let's test the two possible values for $p$:

Case 1: Assume $p = 1.25$

Substitute $p = 1.25$ into the equation for $k$:

$$k = 55 - 48 \times 1.25$$

$$k = 55 - 48 \times \frac{5}{4}$$

$$k = 55 - (12 \times 5)$$

$$k = 55 - 60$$

$$k = -5$$

In this case, we get $p = 1.25$ and $k = -5$. This matches one of the options.

Case 2: Assume $p = -1.25$

Substitute $p = -1.25$ into the equation for $k$:

$$k = 55 - 48 \times (-1.25)$$

$$k = 55 - 48 \times (-\frac{5}{4})$$

$$k = 55 - (-12 \times 5)$$

$$k = 55 - (-60)$$

$$k = 55 + 60$$

$$k = 115$$

In this case, we get $p = -1.25$ and $k = 115$. This does not match any of the options provided.

Conclusion

Based on the calculations, the values that satisfy both the standard deviation and mean transformation properties are $p = 1.25$ and $k = -5$.