Question

1. the following are six lap times recorded in ricky bobbys first car race.

a) calculate the sample standard deviation for the lap times. show your work. 6 marks

lap	time (min)	$(x - \bar{x})^2$
2	12.87	109.7444
3	10.08	87.9844
4	14.24	70.2244
5	11.32	54.4644
6	12.65	40.3044

$s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}$

b) during his second race, rickys lap times had a standard deviation of 5.93 minutes. during which race (first or second) were rickys lap times more consistent? how do you know? 2 marks

1. an interview committee graded 2 candidates for a management position as shown below. the scores had a maximum of 5, and the weightings are out of 3.

a) calculate the weighted mean score for each candidate. 4 marks

criteria	greg	tara	weighting
experience	2	4	3
interpersonal skills	3	5	2
first interview	4	3	1

$\bar{x}_w = \frac{\sum x_i w_i}{\sum w_i}$

Explanation:

Part 2a: Calculate Sample Standard Deviation

Step 1: Find the Mean (\(\bar{x}\))

First, sum the lap times:
\(13.12 + 12.87 + 10.08 + 14.24 + 11.32 + 12.65 = 74.28\)
Then, divide by \(n = 6\):
\(\bar{x} = \frac{74.28}{6} = 12.38\)

Step 2: Verify \(\sum (x - \bar{x})^2\)

From the table, sum the \((x - \bar{x})^2\) values:
\(146.8944 + 109.7444 + 87.9844 + 70.2244 + 54.4644 + 40.3044 = 509.6164\) (Note: The original 508.0164 was likely a typo.)

Step 3: Apply Sample Standard Deviation Formula

The formula is \(s = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}\).
Substitute \(\sum (x - \bar{x})^2 = 509.6164\) and \(n - 1 = 5\):
\(s = \sqrt{\frac{509.6164}{5}} = \sqrt{101.92328} \approx 10.096\)

Part 2b: Consistency of Lap Times

Consistency is determined by standard deviation: smaller \(s\) means less variation (more consistency). The first race has \(s \approx 10.096\) minutes, and the second has \(s = 5.93\) minutes. Since \(5.93 < 10.096\), the second race’s lap times are more consistent.

Part 3a: Weighted Mean for Candidates

The weighted mean formula is \(\bar{x}_w = \frac{\sum (x_i \cdot w_i)}{\sum w_i}\), where \(x_i\) is the score and \(w_i\) is the weight.

For Greg:

Step 1: Calculate Weighted Scores

• Education: \(5 \cdot 2 = 10\)
• Experience: \(2 \cdot 3 = 6\)
• Interpersonal skills: \(3 \cdot 2 = 6\)
• First interview: \(4 \cdot 1 = 4\)

Step 2: Sum Weighted Scores and Weights

\(\sum (x_i \cdot w_i) = 10 + 6 + 6 + 4 = 26\)
\(\sum w_i = 2 + 3 + 2 + 1 = 8\)

Step 3: Compute Weighted Mean

\(\bar{x}_{w,\text{Greg}} = \frac{26}{8} = 3.25\)

For Tara:

Step 1: Calculate Weighted Scores

• Education: \(2 \cdot 2 = 4\)
• Experience: \(4 \cdot 3 = 12\)
• Interpersonal skills: \(5 \cdot 2 = 10\)
• First interview: \(3 \cdot 1 = 3\)

Step 2: Sum Weighted Scores and Weights

\(\sum (x_i \cdot w_i) = 4 + 12 + 10 + 3 = 29\)
\(\sum w_i = 8\) (same as Greg)

Step 3: Compute Weighted Mean

\(\bar{x}_{w,\text{Tara}} = \frac{29}{8} = 3.625\)

Final Answers:

2a) Sample standard deviation \(\approx \boldsymbol{10.10}\) minutes (or \(10.096\) for precision).
2b) The second race; because its standard deviation (\(5.93\)) is smaller than the first’s (\(\approx 10.10\)).
3a) Greg’s weighted mean: \(\boldsymbol{3.25}\); Tara’s weighted mean: \(\boldsymbol{3.625}\).

Answer:

Consistency is determined by standard deviation: smaller \(s\) means less variation (more consistency). The first race has \(s \approx 10.096\) minutes, and the second has \(s = 5.93\) minutes. Since \(5.93 < 10.096\), the second race’s lap times are more consistent.

Part 3a: Weighted Mean for Candidates

The weighted mean formula is \(\bar{x}_w = \frac{\sum (x_i \cdot w_i)}{\sum w_i}\), where \(x_i\) is the score and \(w_i\) is the weight.

For Greg:

Step 1: Calculate Weighted Scores

• Education: \(5 \cdot 2 = 10\)
• Experience: \(2 \cdot 3 = 6\)
• Interpersonal skills: \(3 \cdot 2 = 6\)
• First interview: \(4 \cdot 1 = 4\)

Step 2: Sum Weighted Scores and Weights

\(\sum (x_i \cdot w_i) = 10 + 6 + 6 + 4 = 26\)
\(\sum w_i = 2 + 3 + 2 + 1 = 8\)

Step 3: Compute Weighted Mean

\(\bar{x}_{w,\text{Greg}} = \frac{26}{8} = 3.25\)

For Tara:

Step 1: Calculate Weighted Scores

• Education: \(2 \cdot 2 = 4\)
• Experience: \(4 \cdot 3 = 12\)
• Interpersonal skills: \(5 \cdot 2 = 10\)
• First interview: \(3 \cdot 1 = 3\)

Step 2: Sum Weighted Scores and Weights

\(\sum (x_i \cdot w_i) = 4 + 12 + 10 + 3 = 29\)
\(\sum w_i = 8\) (same as Greg)

Step 3: Compute Weighted Mean

\(\bar{x}_{w,\text{Tara}} = \frac{29}{8} = 3.625\)

Final Answers:

2a) Sample standard deviation \(\approx \boldsymbol{10.10}\) minutes (or \(10.096\) for precision).
2b) The second race; because its standard deviation (\(5.93\)) is smaller than the first’s (\(\approx 10.10\)).
3a) Greg’s weighted mean: \(\boldsymbol{3.25}\); Tara’s weighted mean: \(\boldsymbol{3.625}\).