Question

in lesson app 1.6, we asked, \have you ever noticed that bags of chips seem to contain lots of air and not enough chips?\ here once again are data on the percent of air in each of 14 popular brands of chips, along with a dotplot:

brand	percent of air	brand	percent of air
cheetos	59	pringles	28
doritos	48	ruffles	50
fritos	19	stacy’s pita chips	50
kettle brand	47	sun chips	41
lays	41	terra	49
lays baked	39	tostitos scoops	34

dotplot with x - axis labeled percent of air from 0 to 60, with dots at various positions

1. find the range of the distribution.

1. calculate and interpret the standard deviation.

1. find the interquartile range. interpret this value.

1. the dotplot suggests that the bag of fritos chips, with only 19% of 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:

Question 1: Find the range of the distribution.

Step1: Identify max and min values

The data for percent of air: 46, 59, 48, 19, 47, 41, 39, 45, 28, 50, 50, 41, 49, 34.
Max value = 59, Min value = 19.

Step2: Calculate range (Max - Min)

Range = \( 59 - 19 = 40 \).

Step1: Find the mean (\(\bar{x}\))

Sum of data: \( 46 + 59 + 48 + 19 + 47 + 41 + 39 + 45 + 28 + 50 + 50 + 41 + 49 + 34 = 606 \)
Number of data points (\(n\)) = 14
Mean: \( \bar{x} = \frac{606}{14} \approx 43.2857 \)

Step2: Calculate squared deviations

For each data point \( x_i \), compute \( (x_i - \bar{x})^2 \):

• \( (46 - 43.2857)^2 \approx 7.3776 \)
• \( (59 - 43.2857)^2 \approx 246.9388 \)
• \( (48 - 43.2857)^2 \approx 22.2347 \)
• \( (19 - 43.2857)^2 \approx 590.7347 \)
• \( (47 - 43.2857)^2 \approx 13.8061 \)
• \( (41 - 43.2857)^2 \approx 5.2235 \)
• \( (39 - 43.2857)^2 \approx 18.3571 \)
• \( (45 - 43.2857)^2 \approx 2.9388 \)
• \( (28 - 43.2857)^2 \approx 233.6735 \)
• \( (50 - 43.2857)^2 \approx 45.1837 \)
• \( (50 - 43.2857)^2 \approx 45.1837 \)
• \( (41 - 43.2857)^2 \approx 5.2235 \)
• \( (49 - 43.2857)^2 \approx 32.6531 \)
• \( (34 - 43.2857)^2 \approx 86.2347 \)

Step3: Sum of squared deviations

Sum: \( 7.3776 + 246.9388 + 22.2347 + 590.7347 + 13.8061 + 5.2235 + 18.3571 + 2.9388 + 233.6735 + 45.1837 + 45.1837 + 5.2235 + 32.6531 + 86.2347 \approx 1351.7143 \)

Step4: Calculate variance (\(s^2\)) and standard deviation (\(s\))

Variance (sample): \( s^2 = \frac{1351.7143}{14 - 1} \approx \frac{1351.7143}{13} \approx 103.978 \)
Standard deviation: \( s = \sqrt{103.978} \approx 10.197 \) (≈10.20)

Interpretation:

The standard deviation measures the average distance of data points from the mean. Here, the percent of air in chip bags deviates from the mean (≈43.29%) by about 10.20% on average.

Step1: Order the data

Ordered data: 19, 28, 34, 39, 41, 41, 45, 46, 47, 48, 49, 50, 50, 59

Step2: Find Q1 (25th percentile) and Q3 (75th percentile)

• \( n = 14 \), so the median is between the 7th and 8th values: \( \text{Median} = \frac{45 + 46}{2} = 45.5 \)
• Lower half (first 7 values: 19, 28, 34, 39, 41, 41, 45): Q1 is the median of the lower half (4th value) = 39
• Upper half (last 7 values: 46, 47, 48, 49, 50, 50, 59): Q3 is the median of the upper half (4th value) = 49

Step3: Calculate IQR (Q3 - Q1)

IQR = \( 49 - 39 = 10 \)

Interpretation:

The interquartile range (IQR) is the range of the middle 50% of the data. Here, the middle 50% of chip bag air percentages spans 10 percentage points (from 39% to 49%).

Answer:

40

Question 2: Calculate and interpret the standard deviation.