Question

calculate the population mean, variance, and standard deviation of seven test scores. remember to round appropriately.
66 71 80 100 68 83 74
\\(\mu = \\)
\\(\sigma = \\)
\\(\sigma^2 = \\)

Explanation:

Step1: Calculate the population mean ($\mu$)

The formula for the population mean is $\mu = \frac{\sum_{i = 1}^{n} x_i}{n}$, where $x_i$ are the individual scores and $n$ is the number of scores.
First, sum the scores: $66 + 71 + 80 + 100 + 68 + 83 + 74 = 66 + 71 = 137; 137 + 80 = 217; 217 + 100 = 317; 317 + 68 = 385; 385 + 83 = 468; 468 + 74 = 542$.
There are $n = 7$ scores. So, $\mu = \frac{542}{7} \approx 77.43$.

Step2: Calculate the population variance ($\sigma^2$)

The formula for population variance is $\sigma^2 = \frac{\sum_{i = 1}^{n} (x_i - \mu)^2}{n}$.
Calculate each $(x_i - \mu)^2$:

• For $x = 66$: $(66 - 77.43)^2 = (-11.43)^2 \approx 130.64$
• For $x = 71$: $(71 - 77.43)^2 = (-6.43)^2 \approx 41.34$
• For $x = 80$: $(80 - 77.43)^2 = (2.57)^2 \approx 6.60$
• For $x = 100$: $(100 - 77.43)^2 = (22.57)^2 \approx 509.40$
• For $x = 68$: $(68 - 77.43)^2 = (-9.43)^2 \approx 88.92$
• For $x = 83$: $(83 - 77.43)^2 = (5.57)^2 \approx 31.02$
• For $x = 74$: $(74 - 77.43)^2 = (-3.43)^2 \approx 11.76$

Sum these squared differences: $130.64 + 41.34 + 6.60 + 509.40 + 88.92 + 31.02 + 11.76 = 130.64 + 41.34 = 171.98; 171.98 + 6.60 = 178.58; 178.58 + 509.40 = 687.98; 687.98 + 88.92 = 776.90; 776.90 + 31.02 = 807.92; 807.92 + 11.76 = 819.68$.
Now, divide by $n = 7$: $\sigma^2 = \frac{819.68}{7} \approx 117.097 \approx 117.10$.

Step3: Calculate the population standard deviation ($\sigma$)

The standard deviation is the square root of the variance, so $\sigma = \sqrt{\sigma^2}$.
$\sigma = \sqrt{117.10} \approx 10.82$.

Answer:

$\mu \approx 77.43$, $\sigma \approx 10.82$, $\sigma^2 \approx 117.10$