Question

1. you previously created a five number summary of the lengths of 23 newborns at the dallas university health center. the measurements, in centimeters, were: 46.4 , 46.9, 47.7, 48.1, 48.5, 48.5, 48.8, 49.0, 49.3, 50.0, 50.1, 50.4, 50.6, 51.1, 51.4, 51.8, 52.4, 52.5, 53.2, 53.8, 54.4, 55.1, 55.9 checksum: 1165.9 determine if it is appropriate to summarize the data with the mean and standard deviation. if it is appropriate, justify your reasoning and find the mean and standard deviation. if it is not appropriate, explain why not, and find the median and iqr. consider the precision of the measurement when giving the result. homework help

Explanation:

Step1: Check Data Distribution

First, we analyze the data for symmetry or skewness. Let's list the data points: 46.4, 46.9, 47.7, 48.1, 48.5, 48.5, 48.8, 49.0, 49.3, 50.0, 50.1, 50.4, 50.6, 51.1, 51.4, 51.8, 52.4, 52.5, 53.2, 53.8, 54.4, 55.1, 55.9.
We can create a stem - and - leaf plot or observe the spread. The data seems to be roughly symmetric? Wait, let's check the number of data points. \(n = 23\), which is odd. The median is the 12th value (since \((23 + 1)/2=12\)). The 12th value is 50.4. Let's check the lower half (first 11 values: 46.4, 46.9, 47.7, 48.1, 48.5, 48.5, 48.8, 49.0, 49.3, 50.0, 50.1) and the upper half (last 11 values: 50.6, 51.1, 51.4, 51.8, 52.4, 52.5, 53.2, 53.8, 54.4, 55.1, 55.9).
Wait, actually, to check for skewness, we can also look at the mean and median. Let's calculate the mean first. The sum of the data is given as 1165.9. The mean \(\bar{x}=\frac{\sum x}{n}=\frac{1165.9}{23}\approx50.69\). The median (12th term) is 50.4. The difference between mean and median is small (\(50.69 - 50.4 = 0.29\)), which suggests that the data is approximately symmetric. Also, there are no extreme outliers (since the data points seem to be in a reasonable range for newborn lengths). So, it is appropriate to use mean and standard deviation.

Step2: Calculate the Mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). We know that \(\sum x=1165.9\) and \(n = 23\). So, \(\bar{x}=\frac{1165.9}{23}\approx50.69\) (rounded to two decimal places).

Step3: Calculate the Standard Deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\) (since this is sample data, we use \(n-1\) in the denominator).
First, we calculate \((x_{i}-\bar{x})^{2}\) for each \(x_{i}\):
- For \(x = 46.4\): \((46.4 - 50.69)^{2}=(- 4.29)^{2}=18.4041\)
- For \(x = 46.9\): \((46.9 - 50.69)^{2}=(-3.79)^{2}=14.3641\)
- For \(x = 47.7\): \((47.7 - 50.69)^{2}=(-2.99)^{2}=8.9401\)
- For \(x = 48.1\): \((48.1 - 50.69)^{2}=(-2.59)^{2}=6.7081\)
- For \(x = 48.5\) (first occurrence): \((48.5 - 50.69)^{2}=(-2.19)^{2}=4.7961\)
- For \(x = 48.5\) (second occurrence): \((48.5 - 50.69)^{2}=4.7961\)
- For \(x = 48.8\): \((48.8 - 50.69)^{2}=(-1.89)^{2}=3.5721\)
- For \(x = 49.0\): \((49.0 - 50.69)^{2}=(-1.69)^{2}=2.8561\)
- For \(x = 49.3\): \((49.3 - 50.69)^{2}=(-1.39)^{2}=1.9321\)
- For \(x = 50.0\): \((50.0 - 50.69)^{2}=(-0.69)^{2}=0.4761\)
- For \(x = 50.1\): \((50.1 - 50.69)^{2}=(-0.59)^{2}=0.3481\)
- For \(x = 50.4\): \((50.4 - 50.69)^{2}=(-0.29)^{2}=0.0841\)
- For \(x = 50.6\): \((50.6 - 50.69)^{2}=(-0.09)^{2}=0.0081\)
- For \(x = 51.1\): \((51.1 - 50.69)^{2}=(0.41)^{2}=0.1681\)
- For \(x = 51.4\): \((51.4 - 50.69)^{2}=(0.71)^{2}=0.5041\)
- For \(x = 51.8\): \((51.8 - 50.69)^{2}=(1.11)^{2}=1.2321\)
- For \(x = 52.4\): \((52.4 - 50.69)^{2}=(1.71)^{2}=2.9241\)
- For \(x = 52.5\): \((52.5 - 50.69)^{2}=(1.81)^{2}=3.2761\)
- For \(x = 53.2\): \((53.2 - 50.69)^{2}=(2.51)^{2}=6.3001\)
- For \(x = 53.8\): \((53.8 - 50.69)^{2}=(3.11)^{2}=9.6721\)
- For \(x = 54.4\): \((54.4 - 50.69)^{2}=(3.71)^{2}=13.7641\)
- For \(x = 55.1\): \((55.1 - 50.69)^{2}=(4.41)^{2}=19.4481\)
- For \(x = 55.9\): \((55.9 - 50.69)^{2}=(5.21)^{2}=27.1441\)

Now, sum up all these squared deviations:
\[ \begin{align*} &18.4041+14.3641 + 8.9401+6.7081+4.7961+4.7961+3.5721+2.8561+1.9321+0.4761+0.3481+0.0841+0.0081+0.1681+0.5041+1.2321+2.9241+3.2761+6.3001+9.6721+13.7641+19.4481+27.1441\\ =&(18.4041+14.3641)+(8.9401+6.7081)+(4.7961+4.7961)+(3.5721+2.8561)+(1.9321+0.4761)+(0.3481+0.0841)+(0.0081+0.1681)+(0.5041+1.2321)+(2.9241+3.2761)+(6.3001+9.6721)+(13.7641+19.4481)+27.1441\\ =&32.7682+15.6482+9.5922+6.4282+2.4082+0.4322+0.1762+1.7362+6.2002+15.9722+33.2122+27.1441\\ =&(32.7682+15.6482)+(9.5922+6.4282)+(2.4082+0.4322)+(0.1762+1.7362)+(6.2002+15.9722)+(33.2122+27.1441)\\ =&48.4164+16.0204+2.8404+1.9124+22.1724+60.3563\\ =&(48.4164+16.0204)+(2.8404+1.9124)+(22.1724+60.3563)\\ =&64.4368+4.7528+82.5287\\ =&69.1896+82.5287\\ =&151.7183 \end{align*} \]

Now, \(n - 1=23 - 1 = 22\). So, the variance \(s^{2}=\frac{151.7183}{22}\approx6.8963\). Then the standard deviation \(s=\sqrt{6.8963}\approx2.626\approx2.63\) (rounded to two decimal places).

Answer:

It is appropriate to summarize the data with the mean and standard deviation because the data appears to be approximately symmetric (the difference between the mean and median is small, and there are no extreme outliers). The mean is approximately \(50.69\) cm and the standard deviation is approximately \(2.63\) cm.