Question

1. the temperatures, in °c, at a seaside resort were recorded at noon over a 10-day period.

19 20 19 17 21 18 19 24 25 28
calculate the mean and standard deviation.

1. john plays golf with his brother joe each month. they keep a note of

their scores.

john	74	73	74	73	71	73	72	75	73	73	72	73
joe	68	74	70	67	80	81	69	68	79	67	70	71

calculate the mean and standard deviation and comment on john and
joe’s performance over the year.

Explanation:

Problem 2: Temperatures at a Seaside Resort

Step 1: Calculate the Mean

The mean (\(\bar{x}\)) is the sum of all values divided by the number of values (\(n = 10\)).
Values: \(19, 20, 19, 17, 21, 18, 19, 24, 25, 28\)
Sum: \(19 + 20 + 19 + 17 + 21 + 18 + 19 + 24 + 25 + 28 = 210\)
Mean: \(\bar{x} = \frac{210}{10} = 21\)

Step 2: Calculate the Standard Deviation

Standard deviation (\(s\)) for a sample (or population, here \(n = 10\)) is \(\sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}\).
First, find \((x_i - \bar{x})^2\) for each \(x_i\):

• \(19 - 21 = -2\), \((-2)^2 = 4\) (appears 3 times: \(4 \times 3 = 12\))
• \(20 - 21 = -1\), \((-1)^2 = 1\) (1 time: \(1\))
• \(17 - 21 = -4\), \((-4)^2 = 16\) (1 time: \(16\))
• \(21 - 21 = 0\), \(0^2 = 0\) (1 time: \(0\))
• \(18 - 21 = -3\), \((-3)^2 = 9\) (1 time: \(9\))
• \(24 - 21 = 3\), \(3^2 = 9\) (1 time: \(9\))
• \(25 - 21 = 4\), \(4^2 = 16\) (1 time: \(16\))
• \(28 - 21 = 7\), \(7^2 = 49\) (1 time: \(49\))

Sum of squared deviations: \(12 + 1 + 16 + 0 + 9 + 9 + 16 + 49 = 112\)
Variance: \(\frac{112}{10} = 11.2\)
Standard deviation: \(\sqrt{11.2} \approx 3.35\)

Problem 3: Golf Scores (John and Joe)

John’s Scores:

Step 1: Calculate the Mean

Values: \(74, 73, 74, 73, 71, 73, 72, 75, 73, 73, 72, 73\) (\(n = 12\))
Sum: \(74 + 73 + 74 + 73 + 71 + 73 + 72 + 75 + 73 + 73 + 72 + 73 = 876\)
Mean: \(\bar{x}_J = \frac{876}{12} = 73\)

Step 2: Calculate the Standard Deviation

Find \((x_i - \bar{x}_J)^2\) for each \(x_i\):

• \(74 - 73 = 1\), \(1^2 = 1\) (2 times: \(1 \times 2 = 2\))
• \(73 - 73 = 0\), \(0^2 = 0\) (7 times: \(0 \times 7 = 0\))
• \(71 - 73 = -2\), \((-2)^2 = 4\) (1 time: \(4\))
• \(72 - 73 = -1\), \((-1)^2 = 1\) (2 times: \(1 \times 2 = 2\))
• \(75 - 73 = 2\), \(2^2 = 4\) (1 time: \(4\))

Sum of squared deviations: \(2 + 0 + 4 + 2 + 4 = 12\)
Variance: \(\frac{12}{12} = 1\)
Standard deviation: \(\sqrt{1} = 1\)

Joe’s Scores:

Step 1: Calculate the Mean

Values: \(68, 74, 70, 67, 80, 81, 69, 68, 79, 67, 70, 71\) (\(n = 12\))
Sum: \(68 + 74 + 70 + 67 + 80 + 81 + 69 + 68 + 79 + 67 + 70 + 71 = 864\)
Mean: \(\bar{x}_E = \frac{864}{12} = 72\)

Step 2: Calculate the Standard Deviation

Find \((x_i - \bar{x}_E)^2\) for each \(x_i\):

• \(68 - 72 = -4\), \((-4)^2 = 16\) (2 times: \(16 \times 2 = 32\))
• \(74 - 72 = 2\), \(2^2 = 4\) (1 time: \(4\))
• \(70 - 72 = -2\), \((-2)^2 = 4\) (2 times: \(4 \times 2 = 8\))
• \(67 - 72 = -5\), \((-5)^2 = 25\) (2 times: \(25 \times 2 = 50\))
• \(80 - 72 = 8\), \(8^2 = 64\) (1 time: \(64\))
• \(81 - 72 = 9\), \(9^2 = 81\) (1 time: \(81\))
• \(69 - 72 = -3\), \((-3)^2 = 9\) (1 time: \(9\))
• \(79 - 72 = 7\), \(7^2 = 49\) (1 time: \(49\))
• \(71 - 72 = -1\), \((-1)^2 = 1\) (1 time: \(1\))

Sum of squared deviations: \(32 + 4 + 8 + 50 + 64 + 81 + 9 + 49 + 1 = 298\)
Variance: \(\frac{298}{12} \approx 24.83\)
Standard deviation: \(\sqrt{24.83} \approx 4.98\)

Performance Comment:

• Mean: John’s mean (\(73\)) is higher than Joe’s (\(72\)), so John scores better on average.
• Standard Deviation: John’s standard deviation (\(1\)) is much lower than Joe’s (\(\approx 4.98\)), so John’s scores are more consistent (less variable) than Joe’s.

Answer:

s:

Problem 2:

• Mean: \(21^\circ\text{C}\)
• Standard Deviation: \(\approx 3.35^\circ\text{C}\)

Problem 3:

John:

• Mean: \(73\)
• Standard Deviation: \(1\)

Joe:

• Mean: \(72\)
• Standard Deviation: \(\approx 4.98\)

Comment:

John has a higher mean (better average score) and lower standard deviation (more consistent scores) than Joe.