Question

1. one population has a mean of $mu = 50$ and a standard deviation of $sigma = 15$, and a different population has a mean of $mu = 50$ and a standard deviation of $sigma = 5$. a. sketch both distributions, labelling $mu$ and $sigma$. b. would a score of $x = 65$ be considered an extreme value (out in the tail) in one of these distributions? explain your answer.

Explanation:

Step1: Sketch the distributions

For the first population with $\mu = 50$ and $\sigma=15$, draw a normal - shaped curve centered at 50. Mark the mean $\mu = 50$ on the x - axis. The standard deviation $\sigma = 15$ determines the spread of the curve. For the second population with $\mu = 50$ and $\sigma = 5$, draw another normal - shaped curve centered at 50. Since $\sigma = 5$ is smaller, this curve will be more peaked and have less spread compared to the first one. Mark $\mu = 50$ on its x - axis too.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{X-\mu}{\sigma}$.
For the first population with $\mu = 50$, $\sigma = 15$ and $X = 65$:
$z_1=\frac{65 - 50}{15}=\frac{15}{15}=1$
For the second population with $\mu = 50$, $\sigma = 5$ and $X = 65$:
$z_2=\frac{65 - 50}{5}=\frac{15}{5}=3$

Step3: Determine extreme values

In a normal distribution, values with $|z|\geq3$ are often considered extreme (out in the tails).
Since $z_1 = 1$ and $|z_1|\lt3$, the score $X = 65$ is not an extreme value in the first population.
Since $z_2 = 3$ and $|z_2| = 3$, the score $X = 65$ is an extreme value in the second population.

Answer:

a. Sketch two normal - shaped curves centered at 50 with different spreads according to $\sigma = 15$ and $\sigma = 5$.
b. The score $X = 65$ is not an extreme value in the population with $\mu = 50$ and $\sigma = 15$ because its z - score is $z = 1$. It is an extreme value in the population with $\mu = 50$ and $\sigma = 5$ because its z - score is $z = 3$.