Let's break down each part of the problem step by step.

Part 2: Percentage of Student Weights Above 84 kg

Given:

• Mean ($\mu$) = 68 kg
• Standard deviation ($\sigma$) = 21.9 kg
• We need to find the percentage of students with weights above 84 kg.

To find this, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$ Where:

• $X = 84$ kg
• $\mu = 68$ kg
• $\sigma = 21.9$ kg

Calculate the Z-score: $Z = \frac{84 - 68}{21.9} = \frac{16}{21.9} \approx 0.73$

Next, we look up the Z-score of 0.73 in the standard normal distribution table, which gives the proportion of students with weights less than 84 kg. The value corresponding to $Z = 0.73$ is approximately 0.7673, meaning 76.73% of students have weights below 84 kg.

Therefore, the percentage of students with weights above 84 kg is: $100\% - 76.73\% = 23.27\%$

Part u: Portion of Student Weights Between 54 kg and 70 kg

We need to find the portion of students with weights between 54 kg and 70 kg.

1. Calculate the Z-score for 54 kg: $Z = \frac{54 - 68}{21.9} = \frac{-14}{21.9} \approx -0.64$ From the Z-table, the value corresponding to $Z = -0.64$ is approximately 0.2611.

2. Calculate the Z-score for 70 kg: $Z = \frac{70 - 68}{21.9} = \frac{2}{21.9} \approx 0.09$ From the Z-table, the value corresponding to $Z = 0.09$ is approximately 0.5359.

To find the portion between 54 kg and 70 kg: $0.5359 - 0.2611 = 0.2748$ So, 27.48% of students have weights between 54 kg and 70 kg.

Part e: Values Between Which 70% of Student Weights Fall

We need to find the two values between which 70% of the student weights fall. This means finding the values of $X_1$ and $X_2$ such that 70% of the data lies between them.

In a normal distribution, 70% of the data corresponds to the middle area, so we want the Z-scores that leave 15% (0.15) of the data in each tail (since $0.7 = 1 - 2 \times 0.15$).

From the Z-table, the Z-scores corresponding to 0.15 in each tail are approximately $\pm 1.04$.

Now, convert these Z-scores back to the original values:

1. For $Z = -1.04$: $X_1 = \mu + Z \cdot \sigma = 68 + (-1.04) \cdot 21.9 \approx 45.22 \text{ kg}$

2. For $Z = 1.04$: $X_2 = \mu + Z \cdot \sigma = 68 + 1.04 \cdot 21.9 \approx 90.78 \text{ kg}$

So, 70% of student weights will fall between 45.22 kg and 90.78 kg.

Summary:

1. Percentage of students above 84 kg: 23.27%
2. Portion of student weights between 54 kg and 70 kg: 27.48%
3. 70% of student weights fall between: 45.22 kg and 90.78 kg

Would you like more details or have any other questions?

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Related Questions:

1. What is the probability that a student weighs less than 54 kg?
2. How would the results change if the standard deviation were smaller?
3. What would be the Z-score for a weight of 100 kg?
4. How does the normal distribution assumption affect the conclusions?
5. What is the weight corresponding to the 95th percentile of this distribution?

Tip:

When working with normal distributions, always remember that Z-scores represent the number of standard deviations away from the mean. This helps in understanding where a specific data point lies within the distribution.