To solve these problems, we use the properties of the normal distribution and the z-score formula:

$z = \frac{x - \mu}{\sigma}$ where:

• $z$ is the z-score,
• $x$ is the raw score,
• $\mu$ is the mean,
• $\sigma$ is the standard deviation.

We use the z-table (or normal distribution table) to find probabilities and critical values for given z-scores.

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(a) Proportion of children with scores above 88.5

1. Calculate the z-score for $x = 88.5$: $z = \frac{88.5 - 96}{15} = \frac{-7.5}{15} = -0.5$

2. Look up $z = -0.5$ in the z-table to find the area to the left of this z-score:

   • From the z-table, the area to the left of $z = -0.5$ is 0.3085.

3. The area to the right (proportion of scores above 88.5) is: $P(x > 88.5) = 1 - P(z < -0.5) = 1 - 0.3085 = 0.6915$

Answer (a): $0.6915$

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(b) Lowest 1% of the distribution

1. Find the z-score corresponding to the lowest 1%:

   • From the z-table, the z-score corresponding to the 1st percentile is approximately -2.33.

2. Convert the z-score to the raw score using the z-score formula: $x = z \cdot \sigma + \mu$ $x = -2.33 \cdot 15 + 96 = -34.95 + 96 = 61.05$

3. Round to the nearest tenth: $x = 61.1$

Answer (b): $61.1$

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(c) Highest 48% of the distribution

1. Find the z-score corresponding to the highest 48%:

   • The highest 48% means the cumulative area to the left is $1 - 0.48 = 0.52$.
   • From the z-table, the z-score corresponding to a cumulative area of 0.52 is approximately 0.05.

2. Convert the z-score to the raw score: $x = z \cdot \sigma + \mu$ $x = 0.05 \cdot 15 + 96 = 0.75 + 96 = 96.75$

3. Round to the nearest tenth: $x = 96.8$

Answer (c): $96.8$

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Let me know if you'd like more details on any step!

Here are 5 related questions for practice:

1. What proportion of scores are below 88.5?
2. What score corresponds to the top 10% of this distribution?
3. What is the z-score for a child scoring 110?
4. How many children out of 1,000 are expected to score above 110?
5. What proportion of scores fall between 80 and 100?

Tip: Always verify the cumulative probability from z-tables when working with normal distributions!