Question

fiona recorded the number of miles she biked each day last week as shown below.
4, 7, 4, 10, 5
the mean is given by m = 6. which equation shows the variance for the number of miles fiona biked last week?
$ s^2 = \frac{(4 - 6)^2 + (7 - 6)^2 + (4 - 6)^2 + (10 - 6)^2 + (5 - 6)^2}{6} $
$ \sigma = \sqrt{\frac{(4 - 6)^2 + (7 - 6)^2 + (4 - 6)^2 + (10 - 6)^2 + (5 - 6)^2}{5}} $
$ s = \sqrt{\frac{(4 - 6)^2 + (7 - 6)^2 + (4 - 6)^2 + (10 - 6)^2 + (5 - 6)^2}{4}} $
$ s^2 = \frac{(4 - 6)^2 + (7 - 6)^2 + (4 - 6)^2 + (10 - 6)^2 + (5 - 6)^2}{5} $

Explanation:

Step1: Recall Variance Formula

Variance for a sample is \( s^2=\frac{\sum (x_i - \bar{x})^2}{n - 1} \), but wait, wait—wait, no, wait. Wait, the problem: first, check the data points. The data is 4,7,4,10,5. So number of data points \( n = 5 \). Wait, but let's check the options. Wait, the first option: \( s^2=\frac{(4 - 6)^2+(7 - 6)^2+(4 - 6)^2+(10 - 6)^2+(5 - 6)^2}{6} \)? No, n is 5. Wait, maybe it's a population variance? Wait, the mean is given as m = 6. Wait, let's check each option:

Option 1: \( s^2=\frac{(4 - 6)^2+(7 - 6)^2+(4 - 6)^2+(10 - 6)^2+(5 - 6)^2}{6} \) – denominator 6? No, n=5. Wait, maybe a typo? Wait, no, wait the other options:

Second option: \( \sigma=\sqrt{\frac{(4 - 6)^2+(7 - 6)^2+(4 - 6)^2+(10 - 6)^2+(5 - 6)^2}{5}} \) – that's standard deviation for population.

Third option: \( s=\sqrt{\frac{(4 - 6)^2+(7 - 6)^2+(4 - 6)^2+(10 - 6)^2+(5 - 6)^2}{4}} \) – sample standard deviation (divide by n-1=4).

But the question is about variance. Wait, variance is the square of standard deviation, and for sample variance, it's \( s^2=\frac{\sum (x_i - \bar{x})^2}{n - 1} \), but wait, the first option has denominator 6? Wait, no, maybe the problem is considering population variance? Wait, the data is 5 points. Wait, let's re-express:

Wait, the formula for variance (population) is \( \sigma^2=\frac{\sum (x_i - \mu)^2}{N} \), where N is population size. Sample variance is \( s^2=\frac{\sum (x_i - \bar{x})^2}{n - 1} \).

Given the data: 4,7,4,10,5. So n=5 (number of days, so sample? Or population? Since it's "last week" – maybe 5 days? Wait, the data is 5 numbers: 4,7,4,10,5. So n=5.

Wait, the first option: \( s^2=\frac{(4 - 6)^2+(7 - 6)^2+(4 - 6)^2+(10 - 6)^2+(5 - 6)^2}{6} \) – denominator 6? No, that's wrong. Wait, maybe a mistake in the option? Wait, no, maybe I miscounted. Wait, the data is 4,7,4,10,5 – that's 5 numbers. So n=5.

Wait, the first option's numerator is sum of squared deviations: (4-6)² + (7-6)² + (4-6)² + (10-6)² + (5-6)² – that's correct. Then denominator: if it's sample variance, should be 4; if population, 5. But the first option has 6. Wait, maybe the original problem had a typo, but let's check the options again. Wait, maybe the first option is correct? Wait, no, 5 data points. Wait, maybe the problem is considering "last week" as 6 days? No, the data is 5 numbers. Wait, maybe the user made a typo, but looking at the options, the first option has \( s^2 \) (variance) with numerator sum of squared deviations, and denominator 6? No, that's odd. Wait, no, maybe I misread. Wait, the first option: \( s^2=\frac{(4 - 6)^2+(7 - 6)^2+(4 - 6)^2+(10 - 6)^2+(5 - 6)^2}{6} \) – no, 5 terms, denominator 6? That can't be. Wait, maybe the data was supposed to be 6 days? But the given data is 5. Wait, maybe the first option is correct despite the denominator? No, that doesn't make sense. Wait, maybe the options are:

Wait, the first option: numerator is sum of (x - mean)² for each data point, which is correct. Denominator: if it's sample variance, n-1=4; population, n=5. But the first option has 6. Wait, maybe the problem is wrong, but among the options, the first option is the only one with \( s^2 \) (variance) and the correct numerator. Wait, maybe the denominator is a typo, but let's check the other options:

Second option: \( \sigma=\sqrt{...} \) – that's standard deviation, not variance.

Third option: \( s=\sqrt{...} \) – standard deviation, not variance.

So the first option is the only one with variance ( \( s^2 \) ) and the correct numerator (sum of squared deviations from mean). Even if the denominator is 6 (which is wrong), but maybe it's a mistake, but among the options, the first one is the variance formula (since it's \( s^2 \)) with the correct numerator.

Wait, maybe the original problem had 6 data points, but the given data is 5. But based on the options, the first option is the variance ( \( s^2 \) ) with the sum of (x - mean)², so that's the variance formula (even if denominator is wrong, but maybe a typo). So the first option is the one showing variance.

Step2: Identify the Correct Option

The variance formula for a sample (or population) is the average of squared deviations from the mean. The first option is \( s^2 = \frac{\sum (x_i - m)^2}{n} \) (or n-1), but here the numerator is correct (sum of (4-6)², (7-6)², etc.), and it's expressed as \( s^2 \), which is variance. The other options are standard deviation (with square root), so the first option is the variance equation.

Answer:

\( s^2 = \frac{(4 - 6)^2 + (7 - 6)^2 + (4 - 6)^2 + (10 - 6)^2 + (5 - 6)^2}{6} \) (the first option)