Question

your turn lesson 1.7
why are there so few chips in the bag?
in your turn lesson 1.6, we asked, have you ever noticed that bags of chips seem to contain lots of air and not enough chips? here again are the data on the percentage of air in each of 14 popular brands of chips, along with a dot - plot.

1. find the range of the distribution.
2. calculate and interpret the standard deviation.
3. find the interquartile range. interpret this value.
4. the dot - plot suggests that the bag of fritos chips, with only 19% air, is a possible outlier. recalculate the range, standard deviation, and interquartile range for the other 13 bags of chips. compare these values with the ones you obtained in questions 1 through 3. explain why each result makes sense.

Explanation:

Step1: Find the range for question 1

Range = Max - Min. From the data, Max = 59 (Cheetos) and Min = 19 (Fritos). So Range = 59 - 19 = 40.

Step2: Calculate mean for standard - deviation in question 2

Let the data points be \(x_1,x_2,\cdots,x_{14}\). Mean \(\bar{x}=\frac{\sum_{i = 1}^{14}x_i}{14}\). \(\sum_{i=1}^{14}x_i=46 + 59+48+19+47+41+19+45+28+50+30+45+49+34 = 531\). \(\bar{x}=\frac{531}{14}\approx37.93\).

Step3: Calculate squared - differences for standard - deviation

\((x_1-\bar{x})^2,(x_2 - \bar{x})^2,\cdots\). For example, for \(x_1 = 46\), \((46 - 37.93)^2=(8.07)^2 = 65.1249\). Do this for all data points, sum them up \(\sum_{i = 1}^{14}(x_i-\bar{x})^2\). After calculation, \(\sum_{i = 1}^{14}(x_i-\bar{x})^2 = 1794.9286\).

Step4: Calculate standard - deviation

Standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{14}(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{1794.9286}{13}}\approx11.77\).

Step5: Find quartiles for inter - quartile range in question 3

First, order the data: 19, 19, 28, 30, 34, 41, 45, 45, 46, 47, 48, 49, 50, 59. \(n = 14\). The median (second - quartile \(Q_2\)) is the average of the 7th and 8th ordered values, \(Q_2=\frac{45 + 45}{2}=45\). The first quartile \(Q_1\) is the median of the lower half (19, 19, 28, 30, 34, 41, 45), \(Q_1 = 30\). The third quartile \(Q_3\) is the median of the upper half (45, 46, 47, 48, 49, 50, 59), \(Q_3 = 48\). Inter - quartile range \(IQR=Q_3 - Q_1=48 - 30 = 18\).

Step6: Recalculate for 13 data points in question 4

Remove the value 19 (Fritos). New data set: 19, 28, 30, 34, 41, 45, 45, 46, 47, 48, 49, 50, 59. Max = 59, Min = 19, new range = 59 - 19 = 40 (same as before). Calculate new mean \(\bar{x}_{new}=\frac{19+28+30+34+41+45+45+46+47+48+49+50+59}{13}\approx40.69\). Calculate new \(\sum_{i = 1}^{13}(x_i-\bar{x}_{new})^2\), new standard deviation \(s_{new}=\sqrt{\frac{\sum_{i = 1}^{13}(x_i-\bar{x}_{new})^2}{12}}\approx10.94\). For new \(IQR\), order data, find \(Q_1\), \(Q_2\), \(Q_3\). New \(Q_1 = 34\), \(Q_3 = 48\), new \(IQR = 48 - 34=14\). The range remained the same because the maximum and minimum values didn't change. The standard deviation decreased because the outlier (19) was removed, making the data more clustered. The inter - quartile range decreased because the position of quartiles changed with the removal of the outlier.

Answer:

1. Range: 40
2. Standard deviation: approximately 11.77
3. Inter - quartile range: 18
4. New range: 40, new standard deviation: approximately 10.94, new inter - quartile range: 14.