Question

sixteen students each measured the weight of 25 pennies, in grams. the list shows their measurements. 60, 62, 65, 59, 63, 63, 63, 62, 64, 62, 60, 61, 66, 64, 61, 65 in each graph below, the mean is indicated by a solid vertical line segment while the standard deviations from the mean are indicated by dotted vertical line segments. which graph represents the distribution of weights?

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here, \(n = 16\) and \(x=\{60,62,65,59,63,63,63,62,64,62,60,61,66,64,61,65\}\)
\(\sum_{i=1}^{16}x_{i}=60 + 62+65+59+63+63+63+62+64+62+60+61+66+64+61+65\)
\(\sum_{i = 1}^{16}x_{i}=1000\)
\(\bar{x}=\frac{1000}{16}=62.5\)

Step2: Calculate the standard deviation

The formula for the standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(60 - 62.5)^{2}=(- 2.5)^{2}=6.25\)
\((x_{2}-\bar{x})^{2}=(62 - 62.5)^{2}=(-0.5)^{2}=0.25\)
\((x_{3}-\bar{x})^{2}=(65 - 62.5)^{2}=(2.5)^{2}=6.25\)
\((x_{4}-\bar{x})^{2}=(59 - 62.5)^{2}=(-3.5)^{2}=12.25\)
\((x_{5}-\bar{x})^{2}=(63 - 62.5)^{2}=(0.5)^{2}=0.25\)
\((x_{6}-\bar{x})^{2}=(63 - 62.5)^{2}=(0.5)^{2}=0.25\)
\((x_{7}-\bar{x})^{2}=(63 - 62.5)^{2}=(0.5)^{2}=0.25\)
\((x_{8}-\bar{x})^{2}=(62 - 62.5)^{2}=(-0.5)^{2}=0.25\)
\((x_{9}-\bar{x})^{2}=(64 - 62.5)^{2}=(1.5)^{2}=2.25\)
\((x_{10}-\bar{x})^{2}=(62 - 62.5)^{2}=(-0.5)^{2}=0.25\)
\((x_{11}-\bar{x})^{2}=(60 - 62.5)^{2}=(-2.5)^{2}=6.25\)
\((x_{12}-\bar{x})^{2}=(61 - 62.5)^{2}=(-1.5)^{2}=2.25\)
\((x_{13}-\bar{x})^{2}=(66 - 62.5)^{2}=(3.5)^{2}=12.25\)
\((x_{14}-\bar{x})^{2}=(64 - 62.5)^{2}=(1.5)^{2}=2.25\)
\((x_{15}-\bar{x})^{2}=(61 - 62.5)^{2}=(-1.5)^{2}=2.25\)
\((x_{16}-\bar{x})^{2}=(65 - 62.5)^{2}=(2.5)^{2}=6.25\)

\(\sum_{i = 1}^{16}(x_{i}-\bar{x})^{2}=6.25+0.25 + 6.25+12.25+0.25+0.25+0.25+0.25+2.25+0.25+6.25+2.25+12.25+2.25+2.25+6.25\)
\(\sum_{i = 1}^{16}(x_{i}-\bar{x})^{2}=60\)

\(s=\sqrt{\frac{60}{15}}=\sqrt{4} = 2\)

We know that in a normal - distribution - like graph (assuming the given graphs are of a symmetric distribution around the mean), we check the distance of the data points from the mean.
The mean is \(62.5\approx62\) (approximate to the nearest whole number for graph - checking purpose as per the scale on the y - axis of the graphs) and the standard deviation \(s = 2\)

Answer:

The graph on the right. Since the mean is approximately \(62\) and the standard deviation is \(2\), most of the data should be clustered around \(62\) (within a few standard deviations). The graph on the right has the solid line (mean) at a value close to \(62\) and the distribution is more concentrated around that value compared to the left - hand graph.