Question

A normal distribution has a mean of µ = 80 with σ = 20. What score separates the highest 40% of the distribution from the rest of the scores?

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Answer to a math question A normal distribution has a mean of µ = 80 with σ = 20. What score separates the highest 40% of the distribution from the rest of the scores?

Miles
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To find the score that separates the highest 40% of the distribution from the rest, you can use the z-score formula and then convert it back to the original scale of the distribution. 1. Find the z-score corresponding to the top 40% of the distribution using the standard normal distribution table or calculator. 2. Once you have the z-score, use it to find the corresponding score in the original distribution. The z-score formula is: $z = \frac{x - \mu}{\sigma}$ Where: - $x$ is the score in the distribution - $\mu$ is the mean of the distribution - $\sigma$ is the standard deviation of the distribution - $z$ is the z-score Given: - $\mu = 80$ - $\sigma = 20$ We need to find the z-score corresponding to the top 40% of the distribution, which is $z_{0.40}$. You can find $z_{0.40}$ using a standard normal distribution table or calculator. Once you have $z_{0.40}$, you can find the corresponding score $x$ using the formula: $x = \mu + z_{0.40} \times \sigma$ Let's do the calculations: 1. Find $z_{0.40}$: Using a standard normal distribution table or calculator, $z_{0.40} \approx 0.25$. 2. Find the corresponding score $x$: $x = 80 + 0.25 \times 20 = 80 + 5 = 85$ So, the score that separates the highest 40% of the distribution from the rest of the scores is approximately 85.
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