Question

1-45. 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 measurements when giving the result. (homework help)

Explanation:

Step1: Check Data Distribution

First, we check if the data is approximately symmetric (no extreme skewness or outliers). The data points are: 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. Let's check the number of data points: \( n = 23 \) (odd). The median is the 12th value (since \( \frac{23 + 1}{2}=12 \)): 50.6. The lower half (first 11 values: positions 1 - 11) and upper half (last 11 values: positions 13 - 23). The lower quartile (\( Q_1 \)) is the median of the lower half: 6th value (since \( \frac{11 + 1}{2}=6 \)) of the lower half: 48.5. The upper quartile (\( Q_3 \)) is the median of the upper half: 6th value (positions 13 + 5 = 18th overall? Wait, lower half: 11 values (indices 0 - 10 if 0 - based), so median of lower half (0 - 10) is index 5: 48.5. Upper half: indices 12 - 22 (11 values), median is index 12 + 5 = 17 (0 - based) or 18th (1 - based): 52.5. Now, check for outliers using IQR: \( IQR = Q_3 - Q_1 = 52.5 - 48.5 = 4 \). Lower fence: \( Q_1 - 1.5 \times IQR = 48.5 - 6 = 42.5 \), upper fence: \( Q_3 + 1.5 \times IQR = 52.5 + 6 = 58.5 \). All data points are within [42.5, 58.5], so no outliers. Also, the data seems roughly symmetric (mean and median should be close). So it's appropriate to use mean and standard deviation.

Step2: Calculate the Mean

The sum of the data is given as 1165.9. The mean \( \bar{x} = \frac{\sum x}{n} = \frac{1165.9}{23} \). Let's compute that: \( 1165.9 \div 23 = 50.6913... \approx 50.7 \) (considering precision, measurements are to one decimal place, so mean can be to two or one? Wait, original data has one decimal (e.g., 46.4, 46.9), so sum is 1165.9 (one decimal). So mean: \( \frac{1165.9}{23} = 50.6913 \approx 50.7 \) (or more precise: 50.69).

Step3: Calculate the Standard Deviation

First, we need to compute the variance: \( s^2 = \frac{\sum (x - \bar{x})^2}{n - 1} \). Alternatively, we can use the formula \( \sum x^2 - \frac{(\sum x)^2}{n} \) divided by \( n - 1 \). Let's compute \( \sum x^2 \). Let's square each data point:

46.4² = 2152.96, 46.9² = 2199.61, 47.7² = 2275.29, 48.1² = 2313.61, 48.5² = 2352.25 (two times), 48.8² = 2381.44, 49.0² = 2401, 49.3² = 2430.49, 50.0² = 2500, 50.1² = 2510.01, 50.4² = 2540.16, 50.6² = 2560.36, 51.1² = 2611.21, 51.4² = 2641.96, 51.8² = 2683.24, 52.4² = 2745.76, 52.5² = 2756.25, 53.2² = 2830.24, 53.8² = 2894.44, 54.4² = 2959.36, 55.1² = 3036.01, 55.9² = 3124.81.

Now sum these up: Let's add them step by step.

First, the two 48.5²: 2×2352.25 = 4704.5.

Now sum all squared terms:

2152.96 + 2199.61 = 4352.57; +2275.29 = 6627.86; +2313.61 = 8941.47; +4704.5 = 13645.97; +2381.44 = 16027.41; +2401 = 18428.41; +2430.49 = 20858.9; +2500 = 23358.9; +2510.01 = 25868.91; +2540.16 = 28409.07; +2560.36 = 30969.43; +2611.21 = 33580.64; +2641.96 = 36222.6; +2683.24 = 38905.84; +2745.76 = 41651.6; +2756.25 = 44407.85; +2830.24 = 47238.09; +2894.44 = 50132.53; +2959.36 = 53091.89; +3036.01 = 56127.9; +3124.81 = 59252.71.

So \( \sum x^2 = 59252.71 \).

Now, \( (\sum x)^2 = (1165.9)^2 = 1165.9×1165.9 \). Let's compute that: 1165² = (1100 + 65)² = 1100² + 2×1100×65 + 65² = 1210000 + 143000 + 4225 = 1357225. 0.9² = 0.81, and cross term 2×1165×0.9 = 2097. So (1165.9)² = 1357225 + 2097 + 0.81 = 1359322.81.

Now, \( \sum x^2 - \frac{(\sum x)^2}{n} = 59252.71 - \frac{1359322.81}{23} \). Compute \( \frac{1359322.81}{23} \approx 59100.9917 \). Then, 59252.71 - 59100.9917 = 151.7183.

Now, variance \( s^2 = \frac{151.7183}{23 - 1} = \frac{151.7183}{22} \approx 6.8963 \).

Standard deviation \( s = \sqrt{6.8963} \approx 2.626 \approx 2.63 \) (or 2.6 if we consider precision, but let's check with mean 50.6913).

Alternatively, using the mean we calculated earlier (50.6913), let's verify the sum of (x - x̄)²:

For example, 46.4 - 50.6913 = -4.2913, squared: ~18.415. 46.9 - 50.6913 = -3.7913, squared: ~14.374. Summing all these would be tedious, but the formula using sum of squares is correct. So standard deviation is approximately 2.63.

Answer:

It is appropriate to use mean and standard deviation because the data has no outliers (checked via IQR) and appears roughly symmetric.
Mean: \( \bar{x} = \frac{1165.9}{23} \approx 50.7 \) (or more precisely \( 50.69 \))
Standard Deviation: \( s \approx \sqrt{\frac{59252.71 - \frac{(1165.9)^2}{23}}{22}} \approx 2.63 \)

(If we consider precision of measurements (one decimal place), mean ≈ 50.7 cm, standard deviation ≈ 2.6 cm)