Pregunta
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Use the following distribution to complete parts (a) through (d) below. \[ 3,5,9,12,11 \text { 문 } \] (Round to the nearest hundredth as needed.) b) Multiply each number in the distribution by 2 and compute the mean and the standard deviation of this new distribution. The mean is 16 The standard deviation is 7.75 . (Round to the nearest hundredth as needed.) c) Multiply each number in the original distribution by 4 and compute the mean and the standard deviation of this new distribution. The mean is 32 The standard deviation is 15.48 (Round to the nearest hundredth as needed.) d) The mean and standard deviation of the distribution 2, 4, 5, 5, 6,8 are 5 and 2, respectively. Draw a conclusion about how the mean and standard deviation change when each value in a distribution is multiplied by the same number. and use this conclusion to determine the mean and standard deviation of the distribution \( 6,12,15,15 \), 18, 24 The mean is \( \square \)

Ask by Mccarthy Lyons. in the United States
Mar 16,2025

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The mean is 15 and the standard deviation is 6.

Solución

Given the transformation rule that if each value in a distribution is multiplied by a constant \( k \), then the mean is also multiplied by \( k \) and the standard deviation is multiplied by \( k \), we apply this to the given distributions. The original distribution is \[ 2,4,5,5,6,8. \] It is given that the mean and standard deviation of this distribution are: \[ \text{Mean} = 5 \quad \text{and} \quad \text{Standard Deviation} = 2. \] Now, consider the new distribution: \[ 6,12,15,15,18,24. \] We can observe that each number comes from multiplying the original values by \(3\) (since \(2 \times 3 = 6\), \(4 \times 3 = 12\), etc.). Using the transformation rule: 1. The new mean is: \[ 5 \times 3 = 15. \] 2. The new standard deviation is: \[ 2 \times 3 = 6. \] Thus, the mean for the distribution is \( \boxed{15} \) and the standard deviation is \(6\).

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When each value in a distribution is multiplied by the same number, the mean is also multiplied by that number. For example, in the given problem, when the original values were multiplied by 2, the mean increased from the original mean to double that value. However, the standard deviation is also multiplied by the same factor, thereby stretching the dispersion of the data. At this point, we can apply the same concept to determine the mean and standard deviation for the distribution \(6, 12, 15, 15, 18, 24\). In the new distribution \(6, 12, 15, 15, 18, 24\), the mean is calculated by taking the sum of these numbers (90) and dividing by the total count (6), resulting in a mean of 15. Since we are multiplying each number in the original distribution by 3 (as identified from the transformation process), we can find the standard deviation by tripling the standard deviation of the original distribution, which would be \(\sigma = 2 \times 3 = 6\). Thus, the mean is \(15 \times 3\) which equals \(45\) and the standard deviation is \(6 \times 3 = 18\).

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