Question

the graph shows a distribution of data. what is the variance of the data? 0.0625 0.25 0.5 1.5

Explanation:

Step1: Identify the distribution type

The graph is a normal distribution curve. For a normal distribution, we can estimate the standard deviation (σ) from the spread of the data. The curve is centered at \( \mu = 4 \), and the data seems to spread from about \( 3.5 \) to \( 4.5 \) (within one standard deviation? Wait, actually, looking at the x - axis, the distance from the mean (4) to the points 3.5 and 4.5 is \( 4 - 3.5=0.5 \) and \( 4.5 - 4 = 0.5 \). Wait, no, maybe the standard deviation is related to the width. Wait, in a normal distribution, the variance \( \sigma^{2} \), and standard deviation \( \sigma \). Let's check the options. The options are 0.0625, 0.25, 0.5, 1.5.

Wait, if we look at the x - axis, the distance between 3.5 and 4.5 is 1 unit. But maybe the standard deviation \( \sigma=0.5 \)? No, wait, variance is \( \sigma^{2} \). Wait, let's think again. The mean \( \mu = 4 \). The data points around the mean: the distance from 4 to 3.5 is 0.5, 4 to 4.5 is 0.5. Wait, maybe the standard deviation \( \sigma = 0.5 \)? No, variance is \( \sigma^{2}\). Wait, no, wait 0.25 is \( 0.5^{2}\)? No, \( 0.25 = 0.5\times0.5 \)? Wait, no, \( 0.25=(0.5)^{2}\)? No, \( 0.5^{2}=0.25 \)? Wait, no, \( 0.5\times0.5 = 0.25 \). Wait, but if the standard deviation is 0.5, variance is 0.25? No, wait, no. Wait, maybe the standard deviation is 0.25? Then variance is \( 0.25^{2}=0.0625 \)? No, that's not right. Wait, let's re - examine the graph. The x - axis has marks at 3.5, 4, 4.5. The distance between 3.5 and 4.5 is 1. In a normal distribution, the "width" is related to the standard deviation. Wait, maybe the standard deviation \( \sigma = 0.5 \), then variance \( \sigma^{2}=0.25 \)? No, wait, no. Wait, the options: 0.0625 is \( 0.25^{2}\), 0.25 is \( 0.5^{2}\), 0.5 is just 0.5, 1.5 is 1.5.

Wait, let's think about the normal distribution. The mean is 4. The data is symmetric around 4. The points 3.5 and 4.5 are one standard deviation away? Wait, no, if the standard deviation is 0.5, then the range within one standard deviation is \( \mu-\sigma=4 - 0.5 = 3.5 \) and \( \mu+\sigma=4 + 0.5 = 4.5 \). So \( \sigma = 0.5 \), then variance \( \sigma^{2}=0.5^{2}=0.25 \)? No, wait, \( 0.5^{2}=0.25 \)? Wait, no, \( 0.5\times0.5 = 0.25 \). Wait, but 0.25 is one of the options. Wait, no, wait, maybe I made a mistake. Wait, if the standard deviation is 0.25, then variance is \( 0.25^{2}=0.0625 \). Wait, let's check the x - axis. The distance from 4 to 3.5 is 0.5, so if that's two standard deviations? No, that doesn't make sense. Wait, maybe the graph is a normal distribution with mean 4 and standard deviation 0.5. Then variance is \( 0.5^{2}=0.25 \)? No, \( 0.5^{2}=0.25 \)? Wait, no, \( 0.5\times0.5 = 0.25 \). Wait, but 0.25 is an option. Wait, no, wait, the correct variance: let's recall that variance is the square of the standard deviation. If the standard deviation is 0.5, variance is 0.25? No, \( 0.5^{2}=0.25 \), yes. Wait, but maybe the standard deviation is 0.25, then variance is 0.0625. Wait, the x - axis: from 3.5 to 4.5 is 1 unit. If that's 4 standard deviations (from \( \mu - 2\sigma \) to \( \mu+2\sigma \)), then \( 2\sigma=0.5 \), so \( \sigma = 0.25 \), then variance \( \sigma^{2}=0.0625 \). Ah, that makes sense. Because in a normal distribution, about 95% of the data is within \( \mu\pm2\sigma \). So if the data is within 3.5 (4 - 0.5) and 4.5 (4 + 0.5), then \( 2\sigma=0.5 \), so \( \sigma = 0.25 \), and variance \( \sigma^{2}=0.25^{2}=0.0625 \).

Step2: Calculate the variance

We found that the standard deviation \( \sigma = 0.25 \). The formula for variance is \( \text{Variance}=\sigma^{2} \). So \( \text{Variance}=(0.25)^{2}=0.0625 \).

Answer:

0.0625