Question

1. the following set of data lists (in order) the body temperature (in degree fahrenheit) of the 24 heathy kids at daycare. is this distribution normal, approximately normal, or not normal? show all the work necessary for the proof.

97.2 97.8 98.0 98.2 98.3 98.3 98.5 98.6 98.6 98.7 98.7 98.7
98.8 98.8 98.8 98.8 98.9 99.0 99.1 99.4 99.6 99.7 99.7 99.9

1. the following data represents the number of hours spent on homework per week by a sample of 25 students in a high school.

3, 7, 5, 8, 6, 7, 10, 9, 5, 6, 4, 11, 7, 12, 8, 10, 6, 9, 13, 7, 6, 5, 15, 7, 9
hours spent doing homework
(histogram with frequency on y - axis and hours on x - axis: x - axis labels 3.00, 5.50, 8.00, 10.50, 13.00, 15.50; y - axis labels 0, 2, 4, 6, 8, 10)
using the data and histogram provided, answer the following questions:

Explanation:

Problem 2 (Body Temperature Distribution)

Step 1: Calculate Mean

First, sum all the temperatures. Let's list the data: 97.2, 97.8, 98.0, 98.2, 98.3, 98.3, 98.5, 98.6, 98.6, 98.7, 98.7, 98.7, 98.8, 98.8, 98.8, 98.8, 98.9, 99.0, 99.1, 99.4, 99.6, 99.7, 99.7, 99.9.
Sum = \(97.2 + 97.8 + 98.0 + 98.2 + 98.3 + 98.3 + 98.5 + 98.6 + 98.6 + 98.7 + 98.7 + 98.7 + 98.8 + 98.8 + 98.8 + 98.8 + 98.9 + 99.0 + 99.1 + 99.4 + 99.6 + 99.7 + 99.7 + 99.9\)
Calculate step-by-step:

• First 12 terms: \(97.2+97.8=195\); \(195+98.0=293\); \(293+98.2=391.2\); \(391.2+98.3=489.5\); \(489.5+98.3=587.8\); \(587.8+98.5=686.3\); \(686.3+98.6=784.9\); \(784.9+98.6=883.5\); \(883.5+98.7=982.2\); \(982.2+98.7=1080.9\); \(1080.9+98.7=1179.6\)
• Next 12 terms: \(98.8+98.8=197.6\); \(197.6+98.8=296.4\); \(296.4+98.8=395.2\); \(395.2+98.9=494.1\); \(494.1+99.0=593.1\); \(593.1+99.1=692.2\); \(692.2+99.4=791.6\); \(791.6+99.6=891.2\); \(891.2+99.7=990.9\); \(990.9+99.7=1090.6\); \(1090.6+99.9=1190.5\)

Total sum = \(1179.6 + 1190.5 = 2370.1\)
Mean (\(\mu\)) = \(\frac{2370.1}{24} \approx 98.754\)

Step 2: Calculate Standard Deviation (Approximate)

First, find squared deviations from the mean for each data point, sum them, divide by \(n-1\) (sample) or \(n\) (population), then take square root. Let's use population standard deviation (since it's all 24 kids).
Deviations squared:

• \( (97.2 - 98.754)^2 \approx 2.426 \)
• \( (97.8 - 98.754)^2 \approx 0.910 \)
• \( (98.0 - 98.754)^2 \approx 0.568 \)
• \( (98.2 - 98.754)^2 \approx 0.307 \)
• \( (98.3 - 98.754)^2 \approx 0.206 \) (two times: \(2 \times 0.206 = 0.412\))
• \( (98.5 - 98.754)^2 \approx 0.065 \)
• \( (98.6 - 98.754)^2 \approx 0.024 \) (two times: \(2 \times 0.024 = 0.048\))
• \( (98.7 - 98.754)^2 \approx 0.003 \) (three times: \(3 \times 0.003 = 0.009\))
• \( (98.8 - 98.754)^2 \approx 0.002 \) (four times: \(4 \times 0.002 = 0.008\))
• \( (98.9 - 98.754)^2 \approx 0.021 \)
• \( (99.0 - 98.754)^2 \approx 0.060 \)
• \( (99.1 - 98.754)^2 \approx 0.120 \)
• \( (99.4 - 98.754)^2 \approx 0.417 \)
• \( (99.6 - 98.754)^2 \approx 0.715 \)
• \( (99.7 - 98.754)^2 \approx 0.906 \) (two times: \(2 \times 0.906 = 1.812\))
• \( (99.9 - 98.754)^2 \approx 1.310 \)

Sum of squared deviations: \(2.426 + 0.910 + 0.568 + 0.307 + 0.412 + 0.065 + 0.048 + 0.009 + 0.008 + 0.021 + 0.060 + 0.120 + 0.417 + 0.715 + 1.812 + 1.310 \approx 8.138\)
Variance (\(\sigma^2\)) = \(\frac{8.138}{24} \approx 0.339\)
Standard deviation (\(\sigma\)) = \(\sqrt{0.339} \approx 0.582\)

Step 3: Check Normal Distribution Criteria (Empirical Rule)

For a normal distribution, approximately:

• 68% of data within \(\mu \pm \sigma\)
• 95% within \(\mu \pm 2\sigma\)
• 99.7% within \(\mu \pm 3\sigma\)

Calculate intervals:

• \(\mu \pm \sigma\): \(98.754 \pm 0.582\) → (98.172, 99.336)
• \(\mu \pm 2\sigma\): \(98.754 \pm 1.164\) → (97.59, 99.918)
• \(\mu \pm 3\sigma\): \(98.754 \pm 1.746\) → (96.998, 100.5)

Now count data in each interval:

• Total data: 24
• In \(\mu \pm \sigma\): Data from 98.2 to 99.3 (excluding outliers). Let's count: 98.2, 98.3, 98.3, 98.5, 98.6, 98.6, 98.7, 98.7, 98.7, 98.8, 98.8, 98.8, 98.8, 98.9, 99.0, 99.1. That's 16 data points. \( \frac{16}{24} \approx 66.67\% \), close to 68%.
• In \(\mu \pm 2\sigma\): All data except 97.2, 97.8, 99.4, 99.6, 99.7, 99.7, 99.9? Wait no, 97.2 (97.2 > 97.59? No, 97.2 < 97.59), 97.8 (97.8 > 97.59: yes). Wait 97.2 is below 97.59, 97.8 is above 97.59. Let's list all data:

97.2 (below 97.59), 97.8 (above 97.59), then 98.0 to 99.9. So 97.2 is outside \(\mu \pm 2\sigma\), others: 23 data points. \( \frac{23}{24} \approx 95.83\% \), close to 95%.

• In \(\mu…

Step 1: Analyze Histogram Shape

The histogram has bins: 3.00-5.50, 5.50-8.00, 8.00-10.50, 10.50-13.00, 13.00-15.50.
Frequencies (from data: 3,7,5,8,6,7,10,9,5,6,4,11,7,12,8,10,6,9,13,7,6,5,15,7,9 – wait, 25 data points? Wait the problem says 25 students, but data list has 24? Maybe typo. Anyway, the histogram shows:

• 3.00-5.50: frequency ~5
• 5.50-8.00: frequency ~9
• 8.00-10.50: frequency ~7
• 10.50-13.00: frequency ~2
• 13.00-15.50: frequency ~2

The shape is roughly symmetric? Wait no, left side: 5, right tails: 2,2. Wait maybe slightly skewed right, but the peak is in the middle (5.50-8.00, 8.00-10.50), so maybe approximately normal or slightly skewed. But since the problem is about checking normal distribution (similar to problem 2), we can use empirical rule or check symmetry.

Final Answers (Problem 2)

Answer:

The distribution of body temperatures is approximately normal.

(For Problem 3, more specific questions are needed, but based on the histogram and data, if checking normality, it may be approximately normal or slightly skewed. But since Problem 2 is solved here, the key is Problem 2's answer.)