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Data was collected about the weights for 560 students from a certain University. It was found that the
mean weight for that sample data y = 68 kg and standard deviation o = 21.9 kg. After testing the data,
the study shows that the weights of the students are normally distributed.
2) (3 marks) Find the percentage of student weights above 84 kg.
u) (5 marks) What portion of student weights will be between 54 kg and 70 kg.
e) (6 marks) Between what two values will 70% of the student weights fall?

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?

---

### 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.

Explore how to analyze student weights in a normal distribution with mean 68 kg and standard deviation 21.9 kg. Solve for the percentage of weights above 84 kg, the portion between 54 kg and 70 kg, and where 70% of weights fall. Learn key concepts including Z-scores and the Central Limit Theorem.

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