Question

c) 2,6,6,9,11,14 | 82,86,86,89,91,94

1. standard deviation ii for each lettered part, a through c,

examine the two given sets of numbers. without doing any
calculations, decide which set has the larger standard deviation
and explain why. then check by finding the standard deviations
by hand.

a) set 1: 4,7,7,7,10 | set 2: 4,6,7,8,10
b) set 1: 100,140,150,160,200 | set 2: 10,50,60,70,110
c) set 1: 10,16,18,20,22,28 | set 2: 48,56,58,60,62,70

t 23. pizza prices the histogram shows the distribution of the
prices of plain pizza slices (in $) for 156 weeks in dallas, tx.

which summary statistics would you choose to summarize the
center and spread in these data? why?

t 24. neck size the histogram shows the neck sizes (in inches) of
250 men recruited for a health study in utah.

(histogram for neck size)

t 28. golf drives 2015 the display shows distance (in yards) for 155 professional golfers on the men’s pga tour in 2015.

(histogram for golf drives)

a) describe this distribution.
b) approximately what proportion of...

(other questions about movie run time:)
a) you plan to see a movie this week—how long do you expect a typical movie to run?
b) would you be surprised to find a movie that runs 2.5 hours (150 minutes)?
c) which would you expect to be larger: the mean or the median run time for all movies?

Explanation:

Let's solve part (a) of problem 22 (Standard Deviation II) as an example.

Step 1: Recall the concept of standard deviation

Standard deviation measures the spread (variability) of data points around the mean. A larger standard deviation means the data is more spread out.

Step 2: Analyze Set 1 and Set 2 (part a)

• Set 1: \( 4, 7, 7, 7, 10 \)

Most values are clustered around \( 7 \) (three \( 7 \)s), with \( 4 \) and \( 10 \) as outliers.

• Set 2: \( 4, 6, 7, 8, 10 \)

Values are more evenly distributed: \( 4, 6, 7, 8, 10 \) (no repeated values, spread from \( 4 \) to \( 10 \) with intermediate values).

Step 3: Intuitive Comparison

Set 2 has values that are more spread out (less clustering) than Set 1. Thus, Set 2 should have a larger standard deviation.

Step 4: Verify by Calculating Standard Deviation (Formula: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \))

For Set 1:

• Mean \( \bar{x}_1 = \frac{4 + 7 + 7 + 7 + 10}{5} = \frac{35}{5} = 7 \)
• Deviations: \( (4 - 7)^2 = 9 \), \( (7 - 7)^2 = 0 \) (three times), \( (10 - 7)^2 = 9 \)
• Sum of squared deviations: \( 9 + 0 + 0 + 0 + 9 = 18 \)
• Variance: \( \frac{18}{5 - 1} = \frac{18}{4} = 4.5 \)
• Standard deviation: \( s_1 = \sqrt{4.5} \approx 2.12 \)

For Set 2:

• Mean \( \bar{x}_2 = \frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7 \)
• Deviations: \( (4 - 7)^2 = 9 \), \( (6 - 7)^2 = 1 \), \( (7 - 7)^2 = 0 \), \( (8 - 7)^2 = 1 \), \( (10 - 7)^2 = 9 \)
• Sum of squared deviations: \( 9 + 1 + 0 + 1 + 9 = 20 \)
• Variance: \( \frac{20}{5 - 1} = \frac{20}{4} = 5 \)
• Standard deviation: \( s_2 = \sqrt{5} \approx 2.24 \)

Final Answer (for part a)

Set 2 (\( 4, 6, 7, 8, 10 \)) has a larger standard deviation because its values are more spread out (less clustered) than Set 1. The calculated standard deviations confirm this: \( s_1 \approx 2.12 \) and \( s_2 \approx 2.24 \).

Answer:

Let's solve part (a) of problem 22 (Standard Deviation II) as an example.

Step 1: Recall the concept of standard deviation

Standard deviation measures the spread (variability) of data points around the mean. A larger standard deviation means the data is more spread out.

Step 2: Analyze Set 1 and Set 2 (part a)

• Set 1: \( 4, 7, 7, 7, 10 \)

Most values are clustered around \( 7 \) (three \( 7 \)s), with \( 4 \) and \( 10 \) as outliers.

• Set 2: \( 4, 6, 7, 8, 10 \)

Values are more evenly distributed: \( 4, 6, 7, 8, 10 \) (no repeated values, spread from \( 4 \) to \( 10 \) with intermediate values).

Step 3: Intuitive Comparison

Set 2 has values that are more spread out (less clustering) than Set 1. Thus, Set 2 should have a larger standard deviation.

Step 4: Verify by Calculating Standard Deviation (Formula: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \))

For Set 1:

• Mean \( \bar{x}_1 = \frac{4 + 7 + 7 + 7 + 10}{5} = \frac{35}{5} = 7 \)
• Deviations: \( (4 - 7)^2 = 9 \), \( (7 - 7)^2 = 0 \) (three times), \( (10 - 7)^2 = 9 \)
• Sum of squared deviations: \( 9 + 0 + 0 + 0 + 9 = 18 \)
• Variance: \( \frac{18}{5 - 1} = \frac{18}{4} = 4.5 \)
• Standard deviation: \( s_1 = \sqrt{4.5} \approx 2.12 \)

For Set 2:

• Mean \( \bar{x}_2 = \frac{4 + 6 + 7 + 8 + 10}{5} = \frac{35}{5} = 7 \)
• Deviations: \( (4 - 7)^2 = 9 \), \( (6 - 7)^2 = 1 \), \( (7 - 7)^2 = 0 \), \( (8 - 7)^2 = 1 \), \( (10 - 7)^2 = 9 \)
• Sum of squared deviations: \( 9 + 1 + 0 + 1 + 9 = 20 \)
• Variance: \( \frac{20}{5 - 1} = \frac{20}{4} = 5 \)
• Standard deviation: \( s_2 = \sqrt{5} \approx 2.24 \)

Final Answer (for part a)

Set 2 (\( 4, 6, 7, 8, 10 \)) has a larger standard deviation because its values are more spread out (less clustered) than Set 1. The calculated standard deviations confirm this: \( s_1 \approx 2.12 \) and \( s_2 \approx 2.24 \).