Sally and Ron have decided to go back to school. Sally is 28 and Ron is 25. based on their relative position, which of them (Sally or Ron) would be farther away from the average age of their gender group? Show all steps and work (Hint: you will need to calculate z-scores as part of this question, which are the number of stadard deviations from the mean)

To determine whether Sally or Ron is farther away from the average age of their gender group, we need to calculate their respective z-scores. Z-scores represent the number of standard deviations an individual’s data point is away from the mean.

First, let’s assume that we have data on the ages of individuals in the gender group that both Sally and Ron belong to. We’ll call this group the “gender group sample.”

To calculate the z-score, we’ll need to know the mean (average) and standard deviation of the age in the gender group sample. Let’s assume the mean age of the gender group sample is 30 years old, and the standard deviation is 5 years.

1. Calculate the z-score for Sally:

To calculate Sally’s z-score, we’ll use the formula:

Z = (X – μ) / σ

Where:

Z represents the z-score,

X represents Sally’s age,

μ represents the mean age of the gender group sample, and

σ represents the standard deviation of the gender group sample.

Using the given values:

X = 28

μ = 30

σ = 5

Z = (28 – 30) / 5

Z = -2 / 5

Z = -0.4

Therefore, Sally’s z-score is -0.4.

2. Calculate the z-score for Ron:

Using the same formula as above:

X = 25

μ = 30

σ = 5

Z = (25 – 30) / 5

Z = -5 / 5

Z = -1

Therefore, Ron’s z-score is -1.

Now, to determine who is farther away from the average age of their gender group, we need to compare the absolute values of their z-scores.

|Sally’s z-score| = |-0.4| = 0.4

|Ron’s z-score| = |-1| = 1

Since 1 is greater than 0.4, Ron’s z-score indicates he is farther away from the average age of their gender group compared to Sally’s z-score.

In conclusion, based on their relative positions, Ron is farther away from the average age of their gender group than Sally.